Exercise 4.1 - Cubes & Cube Roots RD Sharma Solutions | Mathematics (Maths) Class 8 PDF Download

 Page 1


Number Cube Classification
1 1 Odd
2 8
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
3 27 Odd Notanevennumber
4 64
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
5 125 Odd Notanevennumber
6 216
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
7 343 Odd Notanevennumber
8 512
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
9 729 Odd Notanevennumber
10 1000
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
Question:1
Find the  cubes of the following numbers:
i 7
ii 12
iii 16
iv 21
v 40
vi 55
vii 100
viii 302
ix 301
Solution:
Cube of a number is given by the number raised to the power three.
i Cube of 7 = 7
3
 = 7 ×7 ×7 = 343
ii  Cube of 12 = 12
3
 = 12 ×12 ×12 = 1728
iii Cube of 16 = 16
3
 = 16 ×16 ×16 = 4096
iv Cube of 21 = 21
3
 = 21 ×21 ×21 = 9261
v Cube of 40 = 40
3
 = 40 ×40 ×40 = 64000
vi Cube of 55 = 55
3
 = 55 ×55 ×55 = 166375
vii Cube of 100 = 100
3
 = 100 ×100 ×100 = 1000000
viii Cube of 302 = 302
3
 = 302 ×302 ×302 = 27543608
ix Cube of 301 = 301
3
 = 301 ×301 ×301 = 27270901
Question:2
Write the cubes of all natural numbers between 1 and 10 and verify the following statements:
i Cubes of all odd natural numbers are odd.
ii Cubes of all even natural numbers are even.
Solution:
The cubes of natural numbers between 1 and 10 are listed and classified in the following table.
We can classify all natural numbers as even or odd number; therefore, to check whether the cube of a natural number is even or odd, it is sufficient to check its divisibility by 2.
If the number is divisible by 2, it is an even number, otherwise it will an odd number.
i From the above table, it is evident that cubes of all odd natural numbers are odd.
ii From the above table, it is evident that cubes of all even natural numbers are even.
 
Question:3
Observe the following pattern:
                1
3
 = 1
        1
3
 + 2
3
 = 1 +2
2
1
3
 + 2
3
 + 3
3
 = 1 +2 +3
2
Write the next three rows and calculate the value of 1
3
 + 2
3 
+ 3
3
 + ... + 9
3 
+ 10
3
 by the above pattern.
Solution:
Extend the pattern as follows:
                                   1
3
= 1                            1
3
+2
3
= (1 +2)
2
                     1
3
+2
3
+3
3
= (1 +2 +3)
2
              1
3
+2
3
+3
3
+4
3
= (1 +2 +3 +4)
2
       1
3
+2
3
+3
3
+4
3
+5
3
= (1 +2 +3 +4 +5)
2
1
3
+2
3
+3
3
+
Now, from the above pattern, the required value is given by:
1
3
+2
3
+3
3
+4
3
+5
3
+6
3
+7
3
+8
3
+9
3
+10
3
= (1 +2 +3 +4 +5 +6 +7 +8 +9 +10)
2
= 55
2
= 3025
Thus, the required value is 3025.
Question:4
Write the cubes of 5 natural numbers which are multiples of 3 and verify the followings:
'The cube of a natural number which is a multiple of 3 is a multiple of 27'
Solution:
Five natural numbers, which are multiples of 3, are 3, 6, 9, 12 and 15.
Cubes of these five numbers are:
  3
3
 = 3 ×3 ×3 = 276
3
 = 6 ×6 ×6 = 2169
3
 = 9 ×9 ×9 = 72912
3
 = 12 ×12 ×12 = 172815
3
 = 15 ×15 ×15 = 3375
Now, let us write the cubes as a multiple of 27. We have:
27 = 27 × 1216 = 27 ×8729 = 27 × 271728 = 27 × 643375 = 27 × 125
It is evident that the cubes of the above multiples of 3 could be written as multiples of 27. Thus, it is verified that the cube of a natural number, which is a multiple of 3, is a multiple of
27.
Page 2


Number Cube Classification
1 1 Odd
2 8
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
3 27 Odd Notanevennumber
4 64
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
5 125 Odd Notanevennumber
6 216
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
7 343 Odd Notanevennumber
8 512
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
9 729 Odd Notanevennumber
10 1000
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
Question:1
Find the  cubes of the following numbers:
i 7
ii 12
iii 16
iv 21
v 40
vi 55
vii 100
viii 302
ix 301
Solution:
Cube of a number is given by the number raised to the power three.
i Cube of 7 = 7
3
 = 7 ×7 ×7 = 343
ii  Cube of 12 = 12
3
 = 12 ×12 ×12 = 1728
iii Cube of 16 = 16
3
 = 16 ×16 ×16 = 4096
iv Cube of 21 = 21
3
 = 21 ×21 ×21 = 9261
v Cube of 40 = 40
3
 = 40 ×40 ×40 = 64000
vi Cube of 55 = 55
3
 = 55 ×55 ×55 = 166375
vii Cube of 100 = 100
3
 = 100 ×100 ×100 = 1000000
viii Cube of 302 = 302
3
 = 302 ×302 ×302 = 27543608
ix Cube of 301 = 301
3
 = 301 ×301 ×301 = 27270901
Question:2
Write the cubes of all natural numbers between 1 and 10 and verify the following statements:
i Cubes of all odd natural numbers are odd.
ii Cubes of all even natural numbers are even.
Solution:
The cubes of natural numbers between 1 and 10 are listed and classified in the following table.
We can classify all natural numbers as even or odd number; therefore, to check whether the cube of a natural number is even or odd, it is sufficient to check its divisibility by 2.
If the number is divisible by 2, it is an even number, otherwise it will an odd number.
i From the above table, it is evident that cubes of all odd natural numbers are odd.
ii From the above table, it is evident that cubes of all even natural numbers are even.
 
Question:3
Observe the following pattern:
                1
3
 = 1
        1
3
 + 2
3
 = 1 +2
2
1
3
 + 2
3
 + 3
3
 = 1 +2 +3
2
Write the next three rows and calculate the value of 1
3
 + 2
3 
+ 3
3
 + ... + 9
3 
+ 10
3
 by the above pattern.
Solution:
Extend the pattern as follows:
                                   1
3
= 1                            1
3
+2
3
= (1 +2)
2
                     1
3
+2
3
+3
3
= (1 +2 +3)
2
              1
3
+2
3
+3
3
+4
3
= (1 +2 +3 +4)
2
       1
3
+2
3
+3
3
+4
3
+5
3
= (1 +2 +3 +4 +5)
2
1
3
+2
3
+3
3
+
Now, from the above pattern, the required value is given by:
1
3
+2
3
+3
3
+4
3
+5
3
+6
3
+7
3
+8
3
+9
3
+10
3
= (1 +2 +3 +4 +5 +6 +7 +8 +9 +10)
2
= 55
2
= 3025
Thus, the required value is 3025.
Question:4
Write the cubes of 5 natural numbers which are multiples of 3 and verify the followings:
'The cube of a natural number which is a multiple of 3 is a multiple of 27'
Solution:
Five natural numbers, which are multiples of 3, are 3, 6, 9, 12 and 15.
Cubes of these five numbers are:
  3
3
 = 3 ×3 ×3 = 276
3
 = 6 ×6 ×6 = 2169
3
 = 9 ×9 ×9 = 72912
3
 = 12 ×12 ×12 = 172815
3
 = 15 ×15 ×15 = 3375
Now, let us write the cubes as a multiple of 27. We have:
27 = 27 × 1216 = 27 ×8729 = 27 × 271728 = 27 × 643375 = 27 × 125
It is evident that the cubes of the above multiples of 3 could be written as multiples of 27. Thus, it is verified that the cube of a natural number, which is a multiple of 3, is a multiple of
27.
Question:5
Write the cubes of 5 natural numbers which are of the form 3n + 1 e. g. 4, 7, 10, . . . and verify the following:
'The cube of a natural number of the form 3n + 1 is a natural number of the same form i.e. when divided by 3 it leaves the remainder 1'.
Solution:
Five natural numbers of the form (3n + 1) could be written by choosing n = 1, 2, 3. . . etc.
Let five such numbers be 4, 7, 10, 13, and 16.
The cubes of these five numbers are:
4
3
= 64, 7
3
= 343, 10
3
= 1000, 13
3
= 2197 and 16
3
= 4096
The cubes of the numbers 4, 7, 10, 13, and 16  could expressed as:
    64 = 3 ×21 +1, which is of the form (3n + 1) for n = 21
  343 = 3 ×114 +1, which is of the form (3n + 1) for n = 114
1000 = 3 ×333 +1, which is of the form (3n + 1) for n = 333
2197 = 3 ×732 +1, which is of the form (3n + 1) for n = 732
4096 = 3 ×1365 +1, which is of the form (3n + 1) for n = 1365
The cubes of the numbers 4, 7, 10, 13, and 16 could be expressed as the natural numbers of the form (3n + 1) for some natural number n; therefore, the statement is verified.
Question:6
Write the cubes of 5 natural numbers of the form 3n + 2 i. e. 5, 8, 11, . . . and verify the following:
'The cube of a natural number of the form 3n + 2 is a natural number of the same form i.e. when it is dividend by 3 the remainder is 2'.
Solution:
Five natural numbers of the form (3n + 2) could be written by choosing n = 1, 2, 3. . . etc.
Let five such numbers be 5, 8, 11, 14, and 17.
The cubes of these five numbers are: 
5
3
= 125, 8
3
= 512, 11
3
= 1331, 14
3
= 2744, and 17
3
= 4913.
The cubes of the numbers 5, 8, 11, 14 and 17 could expressed as:
    125 = 3 ×41 +2, which is of the form (3n + 2) for n = 41
  512 = 3 ×170 +2, which is of the form (3n + 2) for n = 170
1331 = 3 ×443 +2, which is of the form (3n + 2) for n = 443
2744 = 3 ×914 +2, which is of the form (3n + 2) for n = 914
4913 = 3 ×1637 +2, which is of the form (3n + 2) for n = 1637
The cubes of the numbers 5, 8, 11, 14, and 17 can be expressed as the natural numbers of the form (3n + 2) for some natural number n. Hence, the statement is verified.
Question:7
Write the cubes of 5 natural numbers of which are multiples of 7 and verify the following:
'The cube of a multiple of 7 is a multiple of 7
3
'.
Solution:
Five multiples of 7 can be written by choosing different values of a natural number n in the expression 7n.
Let the five multiples be 7, 14, 21, 28 and 35.
The cubes of these numbers are: 
7
3
= 343, 14
3
= 2744, 21
3
= 9261, 28
3
= 21952, and 35
3
= 42875
Now, write the above cubes as a multiple of 7
3
. Proceed as follows:
343 = 7
3
×1
2744 = 14
3
= 14 ×14 ×14 = (7 ×2)×(7 ×2)×(7 ×2) = (7 ×7 ×7)×(2 ×2 ×2) = 7
3
×2
3
9261 = 21
3
= 21 ×21 ×21 = (7 ×3)×(7 ×3)×(7 ×3) = (7 ×7 ×7)×(3 ×3 ×3) = 7
3
×3
3
21952 = 28
3
= 28 ×28 ×28 = (7 ×4)×(7 ×4)×(7 ×4) = (7 ×7 ×7)×(4 ×4 ×4) = 7
3
×4
3
42875 = 35
3
= 35 ×35 ×35 = (7 ×5)×(7 ×5)×(7 ×5) = (7 ×7 ×7)×(5 ×5 ×5) = 7
3
×5
3
Hence, the cube of multiple of 7 is a multiple of 7
3
.
Question:8
Which of the following are perfect cubes?
i 64
ii 216
iii 243
iv 1000
v 1728
vi 3087
vii 4608
viii 106480
ix 166375
x 456533
Solution:
i
On factorising 64 into prime factors, we get
64 = 2 ×2 ×2 ×2 ×2 ×2
Group the factors in triples of equal factors as:
64 = {2 ×2 ×2}×{2 ×2 ×2}
It is evident that the prime factors of 64 can be grouped into triples of equal factors and no factor is left over. Therefore, 64 is a perfect cube.
ii
On factorising 216 into prime factors, we get:
216 = 2 ×2 ×2 ×3 ×3 ×3
Group the factors in triples of equal factors as:
216 = {2 ×2 ×2}×{3 ×3 ×3}
It is evident that the prime factors of 216 can be grouped into triples of equal factors and no factor is left over. Therefore, 216 is a perfect cube.
iii
On factorising 243 into prime factors, we get:
243 = 3 ×3 ×3 ×3 ×3
Group the factors in triples of equal factors as:
243 = {3 ×3 ×3}×3 ×3
Page 3


Number Cube Classification
1 1 Odd
2 8
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
3 27 Odd Notanevennumber
4 64
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
5 125 Odd Notanevennumber
6 216
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
7 343 Odd Notanevennumber
8 512
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
9 729 Odd Notanevennumber
10 1000
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
Question:1
Find the  cubes of the following numbers:
i 7
ii 12
iii 16
iv 21
v 40
vi 55
vii 100
viii 302
ix 301
Solution:
Cube of a number is given by the number raised to the power three.
i Cube of 7 = 7
3
 = 7 ×7 ×7 = 343
ii  Cube of 12 = 12
3
 = 12 ×12 ×12 = 1728
iii Cube of 16 = 16
3
 = 16 ×16 ×16 = 4096
iv Cube of 21 = 21
3
 = 21 ×21 ×21 = 9261
v Cube of 40 = 40
3
 = 40 ×40 ×40 = 64000
vi Cube of 55 = 55
3
 = 55 ×55 ×55 = 166375
vii Cube of 100 = 100
3
 = 100 ×100 ×100 = 1000000
viii Cube of 302 = 302
3
 = 302 ×302 ×302 = 27543608
ix Cube of 301 = 301
3
 = 301 ×301 ×301 = 27270901
Question:2
Write the cubes of all natural numbers between 1 and 10 and verify the following statements:
i Cubes of all odd natural numbers are odd.
ii Cubes of all even natural numbers are even.
Solution:
The cubes of natural numbers between 1 and 10 are listed and classified in the following table.
We can classify all natural numbers as even or odd number; therefore, to check whether the cube of a natural number is even or odd, it is sufficient to check its divisibility by 2.
If the number is divisible by 2, it is an even number, otherwise it will an odd number.
i From the above table, it is evident that cubes of all odd natural numbers are odd.
ii From the above table, it is evident that cubes of all even natural numbers are even.
 
Question:3
Observe the following pattern:
                1
3
 = 1
        1
3
 + 2
3
 = 1 +2
2
1
3
 + 2
3
 + 3
3
 = 1 +2 +3
2
Write the next three rows and calculate the value of 1
3
 + 2
3 
+ 3
3
 + ... + 9
3 
+ 10
3
 by the above pattern.
Solution:
Extend the pattern as follows:
                                   1
3
= 1                            1
3
+2
3
= (1 +2)
2
                     1
3
+2
3
+3
3
= (1 +2 +3)
2
              1
3
+2
3
+3
3
+4
3
= (1 +2 +3 +4)
2
       1
3
+2
3
+3
3
+4
3
+5
3
= (1 +2 +3 +4 +5)
2
1
3
+2
3
+3
3
+
Now, from the above pattern, the required value is given by:
1
3
+2
3
+3
3
+4
3
+5
3
+6
3
+7
3
+8
3
+9
3
+10
3
= (1 +2 +3 +4 +5 +6 +7 +8 +9 +10)
2
= 55
2
= 3025
Thus, the required value is 3025.
Question:4
Write the cubes of 5 natural numbers which are multiples of 3 and verify the followings:
'The cube of a natural number which is a multiple of 3 is a multiple of 27'
Solution:
Five natural numbers, which are multiples of 3, are 3, 6, 9, 12 and 15.
Cubes of these five numbers are:
  3
3
 = 3 ×3 ×3 = 276
3
 = 6 ×6 ×6 = 2169
3
 = 9 ×9 ×9 = 72912
3
 = 12 ×12 ×12 = 172815
3
 = 15 ×15 ×15 = 3375
Now, let us write the cubes as a multiple of 27. We have:
27 = 27 × 1216 = 27 ×8729 = 27 × 271728 = 27 × 643375 = 27 × 125
It is evident that the cubes of the above multiples of 3 could be written as multiples of 27. Thus, it is verified that the cube of a natural number, which is a multiple of 3, is a multiple of
27.
Question:5
Write the cubes of 5 natural numbers which are of the form 3n + 1 e. g. 4, 7, 10, . . . and verify the following:
'The cube of a natural number of the form 3n + 1 is a natural number of the same form i.e. when divided by 3 it leaves the remainder 1'.
Solution:
Five natural numbers of the form (3n + 1) could be written by choosing n = 1, 2, 3. . . etc.
Let five such numbers be 4, 7, 10, 13, and 16.
The cubes of these five numbers are:
4
3
= 64, 7
3
= 343, 10
3
= 1000, 13
3
= 2197 and 16
3
= 4096
The cubes of the numbers 4, 7, 10, 13, and 16  could expressed as:
    64 = 3 ×21 +1, which is of the form (3n + 1) for n = 21
  343 = 3 ×114 +1, which is of the form (3n + 1) for n = 114
1000 = 3 ×333 +1, which is of the form (3n + 1) for n = 333
2197 = 3 ×732 +1, which is of the form (3n + 1) for n = 732
4096 = 3 ×1365 +1, which is of the form (3n + 1) for n = 1365
The cubes of the numbers 4, 7, 10, 13, and 16 could be expressed as the natural numbers of the form (3n + 1) for some natural number n; therefore, the statement is verified.
Question:6
Write the cubes of 5 natural numbers of the form 3n + 2 i. e. 5, 8, 11, . . . and verify the following:
'The cube of a natural number of the form 3n + 2 is a natural number of the same form i.e. when it is dividend by 3 the remainder is 2'.
Solution:
Five natural numbers of the form (3n + 2) could be written by choosing n = 1, 2, 3. . . etc.
Let five such numbers be 5, 8, 11, 14, and 17.
The cubes of these five numbers are: 
5
3
= 125, 8
3
= 512, 11
3
= 1331, 14
3
= 2744, and 17
3
= 4913.
The cubes of the numbers 5, 8, 11, 14 and 17 could expressed as:
    125 = 3 ×41 +2, which is of the form (3n + 2) for n = 41
  512 = 3 ×170 +2, which is of the form (3n + 2) for n = 170
1331 = 3 ×443 +2, which is of the form (3n + 2) for n = 443
2744 = 3 ×914 +2, which is of the form (3n + 2) for n = 914
4913 = 3 ×1637 +2, which is of the form (3n + 2) for n = 1637
The cubes of the numbers 5, 8, 11, 14, and 17 can be expressed as the natural numbers of the form (3n + 2) for some natural number n. Hence, the statement is verified.
Question:7
Write the cubes of 5 natural numbers of which are multiples of 7 and verify the following:
'The cube of a multiple of 7 is a multiple of 7
3
'.
Solution:
Five multiples of 7 can be written by choosing different values of a natural number n in the expression 7n.
Let the five multiples be 7, 14, 21, 28 and 35.
The cubes of these numbers are: 
7
3
= 343, 14
3
= 2744, 21
3
= 9261, 28
3
= 21952, and 35
3
= 42875
Now, write the above cubes as a multiple of 7
3
. Proceed as follows:
343 = 7
3
×1
2744 = 14
3
= 14 ×14 ×14 = (7 ×2)×(7 ×2)×(7 ×2) = (7 ×7 ×7)×(2 ×2 ×2) = 7
3
×2
3
9261 = 21
3
= 21 ×21 ×21 = (7 ×3)×(7 ×3)×(7 ×3) = (7 ×7 ×7)×(3 ×3 ×3) = 7
3
×3
3
21952 = 28
3
= 28 ×28 ×28 = (7 ×4)×(7 ×4)×(7 ×4) = (7 ×7 ×7)×(4 ×4 ×4) = 7
3
×4
3
42875 = 35
3
= 35 ×35 ×35 = (7 ×5)×(7 ×5)×(7 ×5) = (7 ×7 ×7)×(5 ×5 ×5) = 7
3
×5
3
Hence, the cube of multiple of 7 is a multiple of 7
3
.
Question:8
Which of the following are perfect cubes?
i 64
ii 216
iii 243
iv 1000
v 1728
vi 3087
vii 4608
viii 106480
ix 166375
x 456533
Solution:
i
On factorising 64 into prime factors, we get
64 = 2 ×2 ×2 ×2 ×2 ×2
Group the factors in triples of equal factors as:
64 = {2 ×2 ×2}×{2 ×2 ×2}
It is evident that the prime factors of 64 can be grouped into triples of equal factors and no factor is left over. Therefore, 64 is a perfect cube.
ii
On factorising 216 into prime factors, we get:
216 = 2 ×2 ×2 ×3 ×3 ×3
Group the factors in triples of equal factors as:
216 = {2 ×2 ×2}×{3 ×3 ×3}
It is evident that the prime factors of 216 can be grouped into triples of equal factors and no factor is left over. Therefore, 216 is a perfect cube.
iii
On factorising 243 into prime factors, we get:
243 = 3 ×3 ×3 ×3 ×3
Group the factors in triples of equal factors as:
243 = {3 ×3 ×3}×3 ×3
It is evident that the prime factors of 243 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is a not perfect cube.
iv
On factorising 1000 into prime factors, we get:
1000 = 2 ×2 ×2 ×5 ×5 ×5
Group the factors in triples of equal factors as:
1000 = {2 ×2 ×2}×{5 ×5 ×5}
It is evident that the prime factors of 1000 can be grouped into triples of equal factors and no factor is left over. Therefore, 1000 is a perfect cube.
v
On factorising 1728 into prime factors, we get:
1728 = 2 ×2 ×2 ×2 ×2 ×2 ×3 ×3 ×3
Group the factors in triples of equal factors as:
1728 = {2 ×2 ×2}×{2 ×2 ×2}×{3 ×3 ×3}
It is evident that the prime factors of 1728 can be grouped into triples of equal factors and no factor is left over. Therefore, 1728 is a perfect cube.
vi
On factorising 3087 into prime factors, we get:
3087 = 3 ×3 ×7 ×7 ×7
Group the factors in triples of equal factors as:
3087 = 3 ×3 ×{7 ×7 ×7}
It is evident that the prime factors of 3087 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is a not perfect cube.
vii
On factorising 4608 into prime factors, we get:
4608 = 2 ×2 ×2 ×2 ×2 ×2 ×2 ×2 ×2 ×3 ×3
Group the factors in triples of equal factors as:
4608 = {2 ×2 ×2}×{2 ×2 ×2}×{2 ×2 ×2}×3 ×3
It is evident that the prime factors of 4608 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 4608 is a not perfect cube.
viii
On factorising 106480 into prime factors, we get:
106480 = 2 ×2 ×2 ×2 ×5 ×11 ×11 ×11
Group the factors in triples of equal factors as:
106480 = {2 ×2 ×2}×2 ×5 ×{11 ×11 ×11}
It is evident that the prime factors of 106480 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 106480 is a not perfect cube.
ix
On factorising 166375 into prime factors, we get:
166375 = 5 ×5 ×5 ×11 ×11 ×11
Group the factors in triples of equal factors as:
166375 = {5 ×5 ×5}×{11 ×11 ×11}
It is evident that the prime factors of 166375 can be grouped into triples of equal factors and no factor is left over. Therefore, 166375 is a perfect cube.
x
On factorising 456533 into prime factors, we get:
456533 = 7 ×7 ×7 ×11 ×11 ×11
Group the factors in triples of equal factors as:
456533 = {7 ×7 ×7}×{11 ×11 ×11}
It is evident that the prime factors of 456533 can be grouped into triples of equal factors and no factor is left over. Therefore, 456533 is a perfect cube.
Question:9
Which of the following are cubes of even natural numbers?
216, 512, 729, 1000, 3375, 13824
Solution:
We know that the cubes of all even natural numbers are even.
The numbers 216, 512, 1000 and 13824 are cubes of even natural numbers.
The numbers 216, 512, 1000 and 13824 are even and it could be verified by divisibility test of 2, i.e., a number is divisible by 2 if it ends with 0, 2, 4, 6 or 8.
Thus, the cubes of even natural numbers are 216, 512, 1000 and 13824.
Question:10
Which of the following are cubes of odd natural numbers?
125, 343, 1728, 4096, 32768, 6859
Solution:
We know that the cubes of all odd natural numbers are odd. The numbers 125, 343, and 6859 are cubes of odd natural numbers.
Any natural numbers could be either even or odd. Therefore, if a natural number is not even, it is odd. Now, the numbers 125, 343 and 6859 are odd 
Itcouldbeverifiedbydivisibilitytestof2, i. e. , anumberisdivisibleby2ifitendswith0, 2, 4, 6or8. None of the three numbers 125, 343 and 6859 are divisible by 2. Therefore, they are not even,
they are odd. The numbers 1728, 4096 and 32768 are even.
Thus, cubes of odd natural numbers are 125, 343 and 6859.
Question:11
What is the smallest number by which the following numbers must be multiplied, so that the products are perfect cubes?
i 675
ii 1323
iii 2560
iv 7803
v 107811
vi 35721
Solution:
i
On factorising 675 into prime factors, we get:
675 = 3 ×3 ×3 ×5 ×5
On grouping the factors in triples of equal factors, we get:
675 = {3 ×3 ×3}×5 ×5
It is evident that the prime factors of 675 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 675 is a not perfect cube. However, if the number is
multiplied by 5, the factors can be grouped into triples of equal factors and no factor will be left over.
Thus, 675 should be multiplied by 5 to make it a perfect cube.
Page 4


Number Cube Classification
1 1 Odd
2 8
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
3 27 Odd Notanevennumber
4 64
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
5 125 Odd Notanevennumber
6 216
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
7 343 Odd Notanevennumber
8 512
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
9 729 Odd Notanevennumber
10 1000
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
Question:1
Find the  cubes of the following numbers:
i 7
ii 12
iii 16
iv 21
v 40
vi 55
vii 100
viii 302
ix 301
Solution:
Cube of a number is given by the number raised to the power three.
i Cube of 7 = 7
3
 = 7 ×7 ×7 = 343
ii  Cube of 12 = 12
3
 = 12 ×12 ×12 = 1728
iii Cube of 16 = 16
3
 = 16 ×16 ×16 = 4096
iv Cube of 21 = 21
3
 = 21 ×21 ×21 = 9261
v Cube of 40 = 40
3
 = 40 ×40 ×40 = 64000
vi Cube of 55 = 55
3
 = 55 ×55 ×55 = 166375
vii Cube of 100 = 100
3
 = 100 ×100 ×100 = 1000000
viii Cube of 302 = 302
3
 = 302 ×302 ×302 = 27543608
ix Cube of 301 = 301
3
 = 301 ×301 ×301 = 27270901
Question:2
Write the cubes of all natural numbers between 1 and 10 and verify the following statements:
i Cubes of all odd natural numbers are odd.
ii Cubes of all even natural numbers are even.
Solution:
The cubes of natural numbers between 1 and 10 are listed and classified in the following table.
We can classify all natural numbers as even or odd number; therefore, to check whether the cube of a natural number is even or odd, it is sufficient to check its divisibility by 2.
If the number is divisible by 2, it is an even number, otherwise it will an odd number.
i From the above table, it is evident that cubes of all odd natural numbers are odd.
ii From the above table, it is evident that cubes of all even natural numbers are even.
 
Question:3
Observe the following pattern:
                1
3
 = 1
        1
3
 + 2
3
 = 1 +2
2
1
3
 + 2
3
 + 3
3
 = 1 +2 +3
2
Write the next three rows and calculate the value of 1
3
 + 2
3 
+ 3
3
 + ... + 9
3 
+ 10
3
 by the above pattern.
Solution:
Extend the pattern as follows:
                                   1
3
= 1                            1
3
+2
3
= (1 +2)
2
                     1
3
+2
3
+3
3
= (1 +2 +3)
2
              1
3
+2
3
+3
3
+4
3
= (1 +2 +3 +4)
2
       1
3
+2
3
+3
3
+4
3
+5
3
= (1 +2 +3 +4 +5)
2
1
3
+2
3
+3
3
+
Now, from the above pattern, the required value is given by:
1
3
+2
3
+3
3
+4
3
+5
3
+6
3
+7
3
+8
3
+9
3
+10
3
= (1 +2 +3 +4 +5 +6 +7 +8 +9 +10)
2
= 55
2
= 3025
Thus, the required value is 3025.
Question:4
Write the cubes of 5 natural numbers which are multiples of 3 and verify the followings:
'The cube of a natural number which is a multiple of 3 is a multiple of 27'
Solution:
Five natural numbers, which are multiples of 3, are 3, 6, 9, 12 and 15.
Cubes of these five numbers are:
  3
3
 = 3 ×3 ×3 = 276
3
 = 6 ×6 ×6 = 2169
3
 = 9 ×9 ×9 = 72912
3
 = 12 ×12 ×12 = 172815
3
 = 15 ×15 ×15 = 3375
Now, let us write the cubes as a multiple of 27. We have:
27 = 27 × 1216 = 27 ×8729 = 27 × 271728 = 27 × 643375 = 27 × 125
It is evident that the cubes of the above multiples of 3 could be written as multiples of 27. Thus, it is verified that the cube of a natural number, which is a multiple of 3, is a multiple of
27.
Question:5
Write the cubes of 5 natural numbers which are of the form 3n + 1 e. g. 4, 7, 10, . . . and verify the following:
'The cube of a natural number of the form 3n + 1 is a natural number of the same form i.e. when divided by 3 it leaves the remainder 1'.
Solution:
Five natural numbers of the form (3n + 1) could be written by choosing n = 1, 2, 3. . . etc.
Let five such numbers be 4, 7, 10, 13, and 16.
The cubes of these five numbers are:
4
3
= 64, 7
3
= 343, 10
3
= 1000, 13
3
= 2197 and 16
3
= 4096
The cubes of the numbers 4, 7, 10, 13, and 16  could expressed as:
    64 = 3 ×21 +1, which is of the form (3n + 1) for n = 21
  343 = 3 ×114 +1, which is of the form (3n + 1) for n = 114
1000 = 3 ×333 +1, which is of the form (3n + 1) for n = 333
2197 = 3 ×732 +1, which is of the form (3n + 1) for n = 732
4096 = 3 ×1365 +1, which is of the form (3n + 1) for n = 1365
The cubes of the numbers 4, 7, 10, 13, and 16 could be expressed as the natural numbers of the form (3n + 1) for some natural number n; therefore, the statement is verified.
Question:6
Write the cubes of 5 natural numbers of the form 3n + 2 i. e. 5, 8, 11, . . . and verify the following:
'The cube of a natural number of the form 3n + 2 is a natural number of the same form i.e. when it is dividend by 3 the remainder is 2'.
Solution:
Five natural numbers of the form (3n + 2) could be written by choosing n = 1, 2, 3. . . etc.
Let five such numbers be 5, 8, 11, 14, and 17.
The cubes of these five numbers are: 
5
3
= 125, 8
3
= 512, 11
3
= 1331, 14
3
= 2744, and 17
3
= 4913.
The cubes of the numbers 5, 8, 11, 14 and 17 could expressed as:
    125 = 3 ×41 +2, which is of the form (3n + 2) for n = 41
  512 = 3 ×170 +2, which is of the form (3n + 2) for n = 170
1331 = 3 ×443 +2, which is of the form (3n + 2) for n = 443
2744 = 3 ×914 +2, which is of the form (3n + 2) for n = 914
4913 = 3 ×1637 +2, which is of the form (3n + 2) for n = 1637
The cubes of the numbers 5, 8, 11, 14, and 17 can be expressed as the natural numbers of the form (3n + 2) for some natural number n. Hence, the statement is verified.
Question:7
Write the cubes of 5 natural numbers of which are multiples of 7 and verify the following:
'The cube of a multiple of 7 is a multiple of 7
3
'.
Solution:
Five multiples of 7 can be written by choosing different values of a natural number n in the expression 7n.
Let the five multiples be 7, 14, 21, 28 and 35.
The cubes of these numbers are: 
7
3
= 343, 14
3
= 2744, 21
3
= 9261, 28
3
= 21952, and 35
3
= 42875
Now, write the above cubes as a multiple of 7
3
. Proceed as follows:
343 = 7
3
×1
2744 = 14
3
= 14 ×14 ×14 = (7 ×2)×(7 ×2)×(7 ×2) = (7 ×7 ×7)×(2 ×2 ×2) = 7
3
×2
3
9261 = 21
3
= 21 ×21 ×21 = (7 ×3)×(7 ×3)×(7 ×3) = (7 ×7 ×7)×(3 ×3 ×3) = 7
3
×3
3
21952 = 28
3
= 28 ×28 ×28 = (7 ×4)×(7 ×4)×(7 ×4) = (7 ×7 ×7)×(4 ×4 ×4) = 7
3
×4
3
42875 = 35
3
= 35 ×35 ×35 = (7 ×5)×(7 ×5)×(7 ×5) = (7 ×7 ×7)×(5 ×5 ×5) = 7
3
×5
3
Hence, the cube of multiple of 7 is a multiple of 7
3
.
Question:8
Which of the following are perfect cubes?
i 64
ii 216
iii 243
iv 1000
v 1728
vi 3087
vii 4608
viii 106480
ix 166375
x 456533
Solution:
i
On factorising 64 into prime factors, we get
64 = 2 ×2 ×2 ×2 ×2 ×2
Group the factors in triples of equal factors as:
64 = {2 ×2 ×2}×{2 ×2 ×2}
It is evident that the prime factors of 64 can be grouped into triples of equal factors and no factor is left over. Therefore, 64 is a perfect cube.
ii
On factorising 216 into prime factors, we get:
216 = 2 ×2 ×2 ×3 ×3 ×3
Group the factors in triples of equal factors as:
216 = {2 ×2 ×2}×{3 ×3 ×3}
It is evident that the prime factors of 216 can be grouped into triples of equal factors and no factor is left over. Therefore, 216 is a perfect cube.
iii
On factorising 243 into prime factors, we get:
243 = 3 ×3 ×3 ×3 ×3
Group the factors in triples of equal factors as:
243 = {3 ×3 ×3}×3 ×3
It is evident that the prime factors of 243 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is a not perfect cube.
iv
On factorising 1000 into prime factors, we get:
1000 = 2 ×2 ×2 ×5 ×5 ×5
Group the factors in triples of equal factors as:
1000 = {2 ×2 ×2}×{5 ×5 ×5}
It is evident that the prime factors of 1000 can be grouped into triples of equal factors and no factor is left over. Therefore, 1000 is a perfect cube.
v
On factorising 1728 into prime factors, we get:
1728 = 2 ×2 ×2 ×2 ×2 ×2 ×3 ×3 ×3
Group the factors in triples of equal factors as:
1728 = {2 ×2 ×2}×{2 ×2 ×2}×{3 ×3 ×3}
It is evident that the prime factors of 1728 can be grouped into triples of equal factors and no factor is left over. Therefore, 1728 is a perfect cube.
vi
On factorising 3087 into prime factors, we get:
3087 = 3 ×3 ×7 ×7 ×7
Group the factors in triples of equal factors as:
3087 = 3 ×3 ×{7 ×7 ×7}
It is evident that the prime factors of 3087 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is a not perfect cube.
vii
On factorising 4608 into prime factors, we get:
4608 = 2 ×2 ×2 ×2 ×2 ×2 ×2 ×2 ×2 ×3 ×3
Group the factors in triples of equal factors as:
4608 = {2 ×2 ×2}×{2 ×2 ×2}×{2 ×2 ×2}×3 ×3
It is evident that the prime factors of 4608 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 4608 is a not perfect cube.
viii
On factorising 106480 into prime factors, we get:
106480 = 2 ×2 ×2 ×2 ×5 ×11 ×11 ×11
Group the factors in triples of equal factors as:
106480 = {2 ×2 ×2}×2 ×5 ×{11 ×11 ×11}
It is evident that the prime factors of 106480 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 106480 is a not perfect cube.
ix
On factorising 166375 into prime factors, we get:
166375 = 5 ×5 ×5 ×11 ×11 ×11
Group the factors in triples of equal factors as:
166375 = {5 ×5 ×5}×{11 ×11 ×11}
It is evident that the prime factors of 166375 can be grouped into triples of equal factors and no factor is left over. Therefore, 166375 is a perfect cube.
x
On factorising 456533 into prime factors, we get:
456533 = 7 ×7 ×7 ×11 ×11 ×11
Group the factors in triples of equal factors as:
456533 = {7 ×7 ×7}×{11 ×11 ×11}
It is evident that the prime factors of 456533 can be grouped into triples of equal factors and no factor is left over. Therefore, 456533 is a perfect cube.
Question:9
Which of the following are cubes of even natural numbers?
216, 512, 729, 1000, 3375, 13824
Solution:
We know that the cubes of all even natural numbers are even.
The numbers 216, 512, 1000 and 13824 are cubes of even natural numbers.
The numbers 216, 512, 1000 and 13824 are even and it could be verified by divisibility test of 2, i.e., a number is divisible by 2 if it ends with 0, 2, 4, 6 or 8.
Thus, the cubes of even natural numbers are 216, 512, 1000 and 13824.
Question:10
Which of the following are cubes of odd natural numbers?
125, 343, 1728, 4096, 32768, 6859
Solution:
We know that the cubes of all odd natural numbers are odd. The numbers 125, 343, and 6859 are cubes of odd natural numbers.
Any natural numbers could be either even or odd. Therefore, if a natural number is not even, it is odd. Now, the numbers 125, 343 and 6859 are odd 
Itcouldbeverifiedbydivisibilitytestof2, i. e. , anumberisdivisibleby2ifitendswith0, 2, 4, 6or8. None of the three numbers 125, 343 and 6859 are divisible by 2. Therefore, they are not even,
they are odd. The numbers 1728, 4096 and 32768 are even.
Thus, cubes of odd natural numbers are 125, 343 and 6859.
Question:11
What is the smallest number by which the following numbers must be multiplied, so that the products are perfect cubes?
i 675
ii 1323
iii 2560
iv 7803
v 107811
vi 35721
Solution:
i
On factorising 675 into prime factors, we get:
675 = 3 ×3 ×3 ×5 ×5
On grouping the factors in triples of equal factors, we get:
675 = {3 ×3 ×3}×5 ×5
It is evident that the prime factors of 675 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 675 is a not perfect cube. However, if the number is
multiplied by 5, the factors can be grouped into triples of equal factors and no factor will be left over.
Thus, 675 should be multiplied by 5 to make it a perfect cube.
ii
On factorising 1323 into prime factors, we get:
1323 = 3 ×3 ×3 ×7 ×7
On grouping the factors in triples of equal factors, we get:
675 = {3 ×3 ×3}×5 ×5
It is evident that the prime factors of 1323 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 1323 is a not perfect cube. However, if the number
is multiplied by 7, the factors can be grouped into triples of equal factors and no factor will be left over.
Thus, 1323 should be multiplied by 7 to make it a perfect cube.
iii)
On factorising 2560 into prime factors, we get:
2560 = 2 ×2 ×2 ×2 ×2 ×2 ×2 ×2 ×2 ×5
On grouping the factors in triples of equal factors, we get:
2560 = {2 ×2 ×2}×{2 ×2 ×2}×{2 ×2 ×2}×5
It is evident that the prime factors of 2560 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 2560 is a not perfect cube. However, if the number is
multiplied by 5 ×5 = 25, the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 2560 should be multiplied by 25 to make it a perfect cube.
iv
On factorising 7803 into prime factors, we get:
7803 = 3 ×3 ×3 ×17 ×17
On grouping the factors in triples of equal factors, we get:
7803 = {3 ×3 ×3}×17 ×17
It is evident that the prime factors of 7803 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 7803 is a not perfect cube. However, if the number is
multiplied by 17, the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 7803 should be multiplied by 17 to make it a perfect cube.
v
On factorising 107811 into prime factors, we get:
107811 = 3 ×3 ×3 ×3 ×11 ×11 ×11
On grouping the factors in triples of equal factors, we get:
107811 = {3 ×3 ×3}×3 ×{11 ×11 ×11}
It is evident that the prime factors of 107811 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 107811 is a not perfect cube. However, if the number
is multiplied by 3 ×3 = 9, the factors be grouped into triples of equal factors such that no factor is left over.
Thus, 107811 should be multiplied by 9 to make it a perfect cube.
vi
On factorising 35721 into prime factors, we get:
35721 = 3 ×3 ×3 ×3 ×3 ×3 ×7 ×7
On grouping the factors in triples of equal factors, we get:
35721 = {3 ×3 ×3}×{3 ×3 ×3}×7 ×7
It is evident that the prime factors of 35721 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 35721 is a not perfect cube. However, if the number is
multiplied by 7, the factors be grouped into triples of equal factors such that no factor is left over.
Thus, 35721 should be multiplied by 7 to make it a perfect cube.
Question:12
By which smallest number must the following numbers be divided so that the quotient is a perfect cube?
i 675
ii 8640
iii 1600
iv 8788
v 7803
vi 107811
vii 35721
viii 243000
Solution:
i
On factorising 675 into prime factors, we get:
675 = 3 ×3 ×3 ×5 ×5
On grouping the factors in triples of equal factors, we get:
675 = {3 ×3 ×3}×5 ×5
It is evident that the prime factors of 675 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 675 is a not perfect cube. However, if the number is
divided by 5 ×5 = 25, the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 675 should be divided by 25 to make it a perfect cube.
ii
On factorising 8640 into prime factors, we get:
8640 = 2 ×2 ×2 ×2 ×2 ×2 ×3 ×3 ×3 ×5
On grouping the factors in triples of equal factors, we get:
8640 = {2 ×2 ×2}×{2 ×2 ×2}×{3 ×3 ×3}×5
It is evident that the prime factors of 8640 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 8640 is a not perfect cube. However, if the number is
divided by 5, the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 8640 should be divided by 5 to make it a perfect cube.
iii
On factorising 1600 into prime factors, we get:
1600 = 2 ×2 ×2 ×2 ×2 ×2 ×5 ×5
On grouping the factors in triples of equal factors, we get:
1600 = {2 ×2 ×2}×{2 ×2 ×2}×5 ×5
It is evident that the prime factors of 1600 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 1600 is a not perfect cube. However, if the number is
divided by (5 ×5 = 25), the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 1600 should be divided by 25 to make it a perfect cube.
iv
Page 5


Number Cube Classification
1 1 Odd
2 8
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
3 27 Odd Notanevennumber
4 64
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
5 125 Odd Notanevennumber
6 216
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
7 343 Odd Notanevennumber
8 512
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
9 729 Odd Notanevennumber
10 1000
Even 
Lastdigitiseven, i. e. , 0, 2, 4, 6, 8
Question:1
Find the  cubes of the following numbers:
i 7
ii 12
iii 16
iv 21
v 40
vi 55
vii 100
viii 302
ix 301
Solution:
Cube of a number is given by the number raised to the power three.
i Cube of 7 = 7
3
 = 7 ×7 ×7 = 343
ii  Cube of 12 = 12
3
 = 12 ×12 ×12 = 1728
iii Cube of 16 = 16
3
 = 16 ×16 ×16 = 4096
iv Cube of 21 = 21
3
 = 21 ×21 ×21 = 9261
v Cube of 40 = 40
3
 = 40 ×40 ×40 = 64000
vi Cube of 55 = 55
3
 = 55 ×55 ×55 = 166375
vii Cube of 100 = 100
3
 = 100 ×100 ×100 = 1000000
viii Cube of 302 = 302
3
 = 302 ×302 ×302 = 27543608
ix Cube of 301 = 301
3
 = 301 ×301 ×301 = 27270901
Question:2
Write the cubes of all natural numbers between 1 and 10 and verify the following statements:
i Cubes of all odd natural numbers are odd.
ii Cubes of all even natural numbers are even.
Solution:
The cubes of natural numbers between 1 and 10 are listed and classified in the following table.
We can classify all natural numbers as even or odd number; therefore, to check whether the cube of a natural number is even or odd, it is sufficient to check its divisibility by 2.
If the number is divisible by 2, it is an even number, otherwise it will an odd number.
i From the above table, it is evident that cubes of all odd natural numbers are odd.
ii From the above table, it is evident that cubes of all even natural numbers are even.
 
Question:3
Observe the following pattern:
                1
3
 = 1
        1
3
 + 2
3
 = 1 +2
2
1
3
 + 2
3
 + 3
3
 = 1 +2 +3
2
Write the next three rows and calculate the value of 1
3
 + 2
3 
+ 3
3
 + ... + 9
3 
+ 10
3
 by the above pattern.
Solution:
Extend the pattern as follows:
                                   1
3
= 1                            1
3
+2
3
= (1 +2)
2
                     1
3
+2
3
+3
3
= (1 +2 +3)
2
              1
3
+2
3
+3
3
+4
3
= (1 +2 +3 +4)
2
       1
3
+2
3
+3
3
+4
3
+5
3
= (1 +2 +3 +4 +5)
2
1
3
+2
3
+3
3
+
Now, from the above pattern, the required value is given by:
1
3
+2
3
+3
3
+4
3
+5
3
+6
3
+7
3
+8
3
+9
3
+10
3
= (1 +2 +3 +4 +5 +6 +7 +8 +9 +10)
2
= 55
2
= 3025
Thus, the required value is 3025.
Question:4
Write the cubes of 5 natural numbers which are multiples of 3 and verify the followings:
'The cube of a natural number which is a multiple of 3 is a multiple of 27'
Solution:
Five natural numbers, which are multiples of 3, are 3, 6, 9, 12 and 15.
Cubes of these five numbers are:
  3
3
 = 3 ×3 ×3 = 276
3
 = 6 ×6 ×6 = 2169
3
 = 9 ×9 ×9 = 72912
3
 = 12 ×12 ×12 = 172815
3
 = 15 ×15 ×15 = 3375
Now, let us write the cubes as a multiple of 27. We have:
27 = 27 × 1216 = 27 ×8729 = 27 × 271728 = 27 × 643375 = 27 × 125
It is evident that the cubes of the above multiples of 3 could be written as multiples of 27. Thus, it is verified that the cube of a natural number, which is a multiple of 3, is a multiple of
27.
Question:5
Write the cubes of 5 natural numbers which are of the form 3n + 1 e. g. 4, 7, 10, . . . and verify the following:
'The cube of a natural number of the form 3n + 1 is a natural number of the same form i.e. when divided by 3 it leaves the remainder 1'.
Solution:
Five natural numbers of the form (3n + 1) could be written by choosing n = 1, 2, 3. . . etc.
Let five such numbers be 4, 7, 10, 13, and 16.
The cubes of these five numbers are:
4
3
= 64, 7
3
= 343, 10
3
= 1000, 13
3
= 2197 and 16
3
= 4096
The cubes of the numbers 4, 7, 10, 13, and 16  could expressed as:
    64 = 3 ×21 +1, which is of the form (3n + 1) for n = 21
  343 = 3 ×114 +1, which is of the form (3n + 1) for n = 114
1000 = 3 ×333 +1, which is of the form (3n + 1) for n = 333
2197 = 3 ×732 +1, which is of the form (3n + 1) for n = 732
4096 = 3 ×1365 +1, which is of the form (3n + 1) for n = 1365
The cubes of the numbers 4, 7, 10, 13, and 16 could be expressed as the natural numbers of the form (3n + 1) for some natural number n; therefore, the statement is verified.
Question:6
Write the cubes of 5 natural numbers of the form 3n + 2 i. e. 5, 8, 11, . . . and verify the following:
'The cube of a natural number of the form 3n + 2 is a natural number of the same form i.e. when it is dividend by 3 the remainder is 2'.
Solution:
Five natural numbers of the form (3n + 2) could be written by choosing n = 1, 2, 3. . . etc.
Let five such numbers be 5, 8, 11, 14, and 17.
The cubes of these five numbers are: 
5
3
= 125, 8
3
= 512, 11
3
= 1331, 14
3
= 2744, and 17
3
= 4913.
The cubes of the numbers 5, 8, 11, 14 and 17 could expressed as:
    125 = 3 ×41 +2, which is of the form (3n + 2) for n = 41
  512 = 3 ×170 +2, which is of the form (3n + 2) for n = 170
1331 = 3 ×443 +2, which is of the form (3n + 2) for n = 443
2744 = 3 ×914 +2, which is of the form (3n + 2) for n = 914
4913 = 3 ×1637 +2, which is of the form (3n + 2) for n = 1637
The cubes of the numbers 5, 8, 11, 14, and 17 can be expressed as the natural numbers of the form (3n + 2) for some natural number n. Hence, the statement is verified.
Question:7
Write the cubes of 5 natural numbers of which are multiples of 7 and verify the following:
'The cube of a multiple of 7 is a multiple of 7
3
'.
Solution:
Five multiples of 7 can be written by choosing different values of a natural number n in the expression 7n.
Let the five multiples be 7, 14, 21, 28 and 35.
The cubes of these numbers are: 
7
3
= 343, 14
3
= 2744, 21
3
= 9261, 28
3
= 21952, and 35
3
= 42875
Now, write the above cubes as a multiple of 7
3
. Proceed as follows:
343 = 7
3
×1
2744 = 14
3
= 14 ×14 ×14 = (7 ×2)×(7 ×2)×(7 ×2) = (7 ×7 ×7)×(2 ×2 ×2) = 7
3
×2
3
9261 = 21
3
= 21 ×21 ×21 = (7 ×3)×(7 ×3)×(7 ×3) = (7 ×7 ×7)×(3 ×3 ×3) = 7
3
×3
3
21952 = 28
3
= 28 ×28 ×28 = (7 ×4)×(7 ×4)×(7 ×4) = (7 ×7 ×7)×(4 ×4 ×4) = 7
3
×4
3
42875 = 35
3
= 35 ×35 ×35 = (7 ×5)×(7 ×5)×(7 ×5) = (7 ×7 ×7)×(5 ×5 ×5) = 7
3
×5
3
Hence, the cube of multiple of 7 is a multiple of 7
3
.
Question:8
Which of the following are perfect cubes?
i 64
ii 216
iii 243
iv 1000
v 1728
vi 3087
vii 4608
viii 106480
ix 166375
x 456533
Solution:
i
On factorising 64 into prime factors, we get
64 = 2 ×2 ×2 ×2 ×2 ×2
Group the factors in triples of equal factors as:
64 = {2 ×2 ×2}×{2 ×2 ×2}
It is evident that the prime factors of 64 can be grouped into triples of equal factors and no factor is left over. Therefore, 64 is a perfect cube.
ii
On factorising 216 into prime factors, we get:
216 = 2 ×2 ×2 ×3 ×3 ×3
Group the factors in triples of equal factors as:
216 = {2 ×2 ×2}×{3 ×3 ×3}
It is evident that the prime factors of 216 can be grouped into triples of equal factors and no factor is left over. Therefore, 216 is a perfect cube.
iii
On factorising 243 into prime factors, we get:
243 = 3 ×3 ×3 ×3 ×3
Group the factors in triples of equal factors as:
243 = {3 ×3 ×3}×3 ×3
It is evident that the prime factors of 243 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is a not perfect cube.
iv
On factorising 1000 into prime factors, we get:
1000 = 2 ×2 ×2 ×5 ×5 ×5
Group the factors in triples of equal factors as:
1000 = {2 ×2 ×2}×{5 ×5 ×5}
It is evident that the prime factors of 1000 can be grouped into triples of equal factors and no factor is left over. Therefore, 1000 is a perfect cube.
v
On factorising 1728 into prime factors, we get:
1728 = 2 ×2 ×2 ×2 ×2 ×2 ×3 ×3 ×3
Group the factors in triples of equal factors as:
1728 = {2 ×2 ×2}×{2 ×2 ×2}×{3 ×3 ×3}
It is evident that the prime factors of 1728 can be grouped into triples of equal factors and no factor is left over. Therefore, 1728 is a perfect cube.
vi
On factorising 3087 into prime factors, we get:
3087 = 3 ×3 ×7 ×7 ×7
Group the factors in triples of equal factors as:
3087 = 3 ×3 ×{7 ×7 ×7}
It is evident that the prime factors of 3087 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is a not perfect cube.
vii
On factorising 4608 into prime factors, we get:
4608 = 2 ×2 ×2 ×2 ×2 ×2 ×2 ×2 ×2 ×3 ×3
Group the factors in triples of equal factors as:
4608 = {2 ×2 ×2}×{2 ×2 ×2}×{2 ×2 ×2}×3 ×3
It is evident that the prime factors of 4608 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 4608 is a not perfect cube.
viii
On factorising 106480 into prime factors, we get:
106480 = 2 ×2 ×2 ×2 ×5 ×11 ×11 ×11
Group the factors in triples of equal factors as:
106480 = {2 ×2 ×2}×2 ×5 ×{11 ×11 ×11}
It is evident that the prime factors of 106480 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 106480 is a not perfect cube.
ix
On factorising 166375 into prime factors, we get:
166375 = 5 ×5 ×5 ×11 ×11 ×11
Group the factors in triples of equal factors as:
166375 = {5 ×5 ×5}×{11 ×11 ×11}
It is evident that the prime factors of 166375 can be grouped into triples of equal factors and no factor is left over. Therefore, 166375 is a perfect cube.
x
On factorising 456533 into prime factors, we get:
456533 = 7 ×7 ×7 ×11 ×11 ×11
Group the factors in triples of equal factors as:
456533 = {7 ×7 ×7}×{11 ×11 ×11}
It is evident that the prime factors of 456533 can be grouped into triples of equal factors and no factor is left over. Therefore, 456533 is a perfect cube.
Question:9
Which of the following are cubes of even natural numbers?
216, 512, 729, 1000, 3375, 13824
Solution:
We know that the cubes of all even natural numbers are even.
The numbers 216, 512, 1000 and 13824 are cubes of even natural numbers.
The numbers 216, 512, 1000 and 13824 are even and it could be verified by divisibility test of 2, i.e., a number is divisible by 2 if it ends with 0, 2, 4, 6 or 8.
Thus, the cubes of even natural numbers are 216, 512, 1000 and 13824.
Question:10
Which of the following are cubes of odd natural numbers?
125, 343, 1728, 4096, 32768, 6859
Solution:
We know that the cubes of all odd natural numbers are odd. The numbers 125, 343, and 6859 are cubes of odd natural numbers.
Any natural numbers could be either even or odd. Therefore, if a natural number is not even, it is odd. Now, the numbers 125, 343 and 6859 are odd 
Itcouldbeverifiedbydivisibilitytestof2, i. e. , anumberisdivisibleby2ifitendswith0, 2, 4, 6or8. None of the three numbers 125, 343 and 6859 are divisible by 2. Therefore, they are not even,
they are odd. The numbers 1728, 4096 and 32768 are even.
Thus, cubes of odd natural numbers are 125, 343 and 6859.
Question:11
What is the smallest number by which the following numbers must be multiplied, so that the products are perfect cubes?
i 675
ii 1323
iii 2560
iv 7803
v 107811
vi 35721
Solution:
i
On factorising 675 into prime factors, we get:
675 = 3 ×3 ×3 ×5 ×5
On grouping the factors in triples of equal factors, we get:
675 = {3 ×3 ×3}×5 ×5
It is evident that the prime factors of 675 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 675 is a not perfect cube. However, if the number is
multiplied by 5, the factors can be grouped into triples of equal factors and no factor will be left over.
Thus, 675 should be multiplied by 5 to make it a perfect cube.
ii
On factorising 1323 into prime factors, we get:
1323 = 3 ×3 ×3 ×7 ×7
On grouping the factors in triples of equal factors, we get:
675 = {3 ×3 ×3}×5 ×5
It is evident that the prime factors of 1323 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 1323 is a not perfect cube. However, if the number
is multiplied by 7, the factors can be grouped into triples of equal factors and no factor will be left over.
Thus, 1323 should be multiplied by 7 to make it a perfect cube.
iii)
On factorising 2560 into prime factors, we get:
2560 = 2 ×2 ×2 ×2 ×2 ×2 ×2 ×2 ×2 ×5
On grouping the factors in triples of equal factors, we get:
2560 = {2 ×2 ×2}×{2 ×2 ×2}×{2 ×2 ×2}×5
It is evident that the prime factors of 2560 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 2560 is a not perfect cube. However, if the number is
multiplied by 5 ×5 = 25, the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 2560 should be multiplied by 25 to make it a perfect cube.
iv
On factorising 7803 into prime factors, we get:
7803 = 3 ×3 ×3 ×17 ×17
On grouping the factors in triples of equal factors, we get:
7803 = {3 ×3 ×3}×17 ×17
It is evident that the prime factors of 7803 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 7803 is a not perfect cube. However, if the number is
multiplied by 17, the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 7803 should be multiplied by 17 to make it a perfect cube.
v
On factorising 107811 into prime factors, we get:
107811 = 3 ×3 ×3 ×3 ×11 ×11 ×11
On grouping the factors in triples of equal factors, we get:
107811 = {3 ×3 ×3}×3 ×{11 ×11 ×11}
It is evident that the prime factors of 107811 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 107811 is a not perfect cube. However, if the number
is multiplied by 3 ×3 = 9, the factors be grouped into triples of equal factors such that no factor is left over.
Thus, 107811 should be multiplied by 9 to make it a perfect cube.
vi
On factorising 35721 into prime factors, we get:
35721 = 3 ×3 ×3 ×3 ×3 ×3 ×7 ×7
On grouping the factors in triples of equal factors, we get:
35721 = {3 ×3 ×3}×{3 ×3 ×3}×7 ×7
It is evident that the prime factors of 35721 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 35721 is a not perfect cube. However, if the number is
multiplied by 7, the factors be grouped into triples of equal factors such that no factor is left over.
Thus, 35721 should be multiplied by 7 to make it a perfect cube.
Question:12
By which smallest number must the following numbers be divided so that the quotient is a perfect cube?
i 675
ii 8640
iii 1600
iv 8788
v 7803
vi 107811
vii 35721
viii 243000
Solution:
i
On factorising 675 into prime factors, we get:
675 = 3 ×3 ×3 ×5 ×5
On grouping the factors in triples of equal factors, we get:
675 = {3 ×3 ×3}×5 ×5
It is evident that the prime factors of 675 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 675 is a not perfect cube. However, if the number is
divided by 5 ×5 = 25, the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 675 should be divided by 25 to make it a perfect cube.
ii
On factorising 8640 into prime factors, we get:
8640 = 2 ×2 ×2 ×2 ×2 ×2 ×3 ×3 ×3 ×5
On grouping the factors in triples of equal factors, we get:
8640 = {2 ×2 ×2}×{2 ×2 ×2}×{3 ×3 ×3}×5
It is evident that the prime factors of 8640 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 8640 is a not perfect cube. However, if the number is
divided by 5, the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 8640 should be divided by 5 to make it a perfect cube.
iii
On factorising 1600 into prime factors, we get:
1600 = 2 ×2 ×2 ×2 ×2 ×2 ×5 ×5
On grouping the factors in triples of equal factors, we get:
1600 = {2 ×2 ×2}×{2 ×2 ×2}×5 ×5
It is evident that the prime factors of 1600 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 1600 is a not perfect cube. However, if the number is
divided by (5 ×5 = 25), the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 1600 should be divided by 25 to make it a perfect cube.
iv
On factorising 8788 into prime factors, we get:
8788 = 2 ×2 ×13 ×13 ×13
On grouping the factors in triples of equal factors, we get:
8788 = 2 ×2 ×{13 ×13 ×13}
It is evident that the prime factors of 8788 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 8788 is a not perfect cube. However, if the number is
divided by (2 ×2 = 4), the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 8788 should be divided by 4 to make it a perfect cube.
v
On factorising 7803 into prime factors, we get:
7803 = 3 ×3 ×3 ×17 ×17
On grouping the factors in triples of equal factors, we get:
7803 = {3 ×3 ×3}×17 ×17
It is evident that the prime factors of 7803 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 7803 is a not perfect cube. However, if the number is
divided by (17 ×17 = 289), the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 7803 should be divided by 289 to  make it a perfect cube.
vi
On factorising 107811 into prime factors, we get:
107811 = 3 ×3 ×3 ×3 ×11 ×11 ×11
On group the factors in triples of equal factors, we get:
107811 = {3 ×3 ×3}×3 ×{11 ×11 ×11}
It is evident that the prime factors of 107811 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 107811 is a not perfect cube. However, if the number
is divided by 3, the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 107811 should be divided by 3 to make it a perfect cube.
vii
On factorising 35721 into prime factors, we get:
35721 = 3 ×3 ×3 ×3 ×3 ×3 ×7 ×7
On grouping the factors in triples of equal factors, we get:
35721 = {3 ×3 ×3}×{3 ×3 ×3}×7 ×7
It is evident that the prime factors of 35721 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 35721 is a not perfect cube. However, if the number is
divided by (7 ×7 = 49), the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 35721 should be divided by 49 to make it a perfect cube.
viii
On factorising 243000 into prime factors, we get:
243000 = 2 ×2 ×2 ×3 ×3 ×3 ×3 ×3 ×5 ×5 ×5
On grouping the factors in triples of equal factors, we get:
243000 = {2 ×2 ×2}×{3 ×3 ×3}×3 ×3 ×{5 ×5 ×5}
It is evident that the prime factors of 243000 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243000 is a not perfect cube. However, if the number
is divided by (3 ×3 = 9), the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 243000 should be divided by 9 to make it a perfect cube.
Question:13
Prove that if a number is trebled then its cube is 27 times the cube of the given number.
Solution:
Let us consider a number n. Then its cube would be n
3
.
If the number n is trebled, i.e., 3n, we get:
(3n)
3
= 3
3
×n
3
= 27n
3
It is evident that the cube of 3n is 27 times of the cube of n.
Hence, the statement is proved.
Question:14
What happens to the cube of a number if the number is multiplied by
i 3?
ii 4?
iii 5?
Solution:
i
Let us consider a number n. Its cube would be n
3
. If n is multiplied by 3, it becomes 3n.
Let us now find the cube of 3n, we get:
(3n)
3
= 3
3
×n
3
= 27n
3
Therefore, the cube of 3n is 27 times of the cube of n.
Thus, if a number is multiplied by 3, its cube is 27 times of the cube of that number.
ii
Let us consider a number n. Its cube would be n
3
. If n is multiplied by 4, it becomes 4n.
Let us now find the cube of 4n, we get:
(4n)
3
= 4
3
×n
3
= 64n
3
Therefore, the cube of 4n is 64 times of the cube of n.
Thus, if a number is multiplied by 4, its cube is 64 times of the cube of that number.
iii
Let us consider a number n. Its cube would be n
3
. If the number n is multiplied by 5, it becomes 5n.
Let us now find the cube of 4n, we get:
(5n)
3
= 5
3
×n
3
= 125n
3
Therefore, the cube of 5n is 125 times of the cube of n.
Thus, if a number is multiplied by 5, its cube is 125 times of the cube of that number.
Question:15
Find the volume of a cube, one face of which has an area of 64 m
2
.
Solution:
Area of a face of cube is given by:
 A = s
2
, where s = Side of the cube