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Class 8 Exam > Class 8 Notes > RD Sharma Solutions for Class 8 Mathematics > RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1)

RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) | RD Sharma Solutions for Class 8 Mathematics PDF Download

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

Answer 1: Cube of a number is given by the number raised to the power three.
(i) $Cube of 7 = 7^{3} = 7 \times 7 \times 7 = 343$
(ii) $Cube of 12 = 12^{3} = 12 \times 12 \times 12 = 1728$
(iii) $Cube of 16 = 16^{3} = 16 \times 16 \times 16 = 4096$
(iv) $Cube of 21 = 21^{3} = 21 \times 21 \times 21 = 9261$
(v) $Cube of 40 = 40^{3} = 40 \times 40 \times 40 = 64000$
(vi) $Cube of 55 = 55^{3} = 55 \times 55 \times 55 = 166375$
(vii) $Cube of 100 = 100^{3} = 100 \times 100 \times 100 = 1000000$
(viii) $Cube of 302 = 302^{3} = 302 \times 302 \times 302 = 27543608$
(ix) $Cube of 301 = 301^{3} = 301 \times 301 \times 301 = 27270901$

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.

Answer 2: 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.

Number	Cube	Classification
1	1	Odd
2	8	Even (Last digit is even, i.e., 0, 2, 4, 6, 8)
3	27	Odd (Not an even number)
4	64	Even (Last digit is even, i.e., 0, 2, 4, 6, 8)
5	125	Odd (Not an even number)
6	216	Even (Last digit is even, i.e., 0, 2, 4, 6, 8)
7	343	Odd (Not an even number)
8	512	Even (Last digit is even, i.e., 0, 2, 4, 6, 8)
9	729	Odd (Not an even number)
10	1000	Even (Last digit is even, i.e., 0, 2, 4, 6, 8)

Question 3: Observe the following pattern:
1³ = 1
1³ + 2³ = (1 + 2)²
1³ + 2³ + 3³ = (1 + 2 + 3)²

Write the next three rows and calculate the value of 1³ + 2³+ 3³ + ... + 9³+ 10³ by the above pattern.

Answer 3: Extend the pattern as follows:

13=1 13+23=(1+2)2 13+23+33=(1+2+3)2 13+23+33+43=(1+2+3+4)2 13+23+33+43+53=(1+2+3+4+5)213+23+33+43+53+63=(1+2+3+4+5+6)2

Now, from the above pattern, the required value is given by:

1³+2³+3³+4³+5³+6³+7³+8³+9³+10³=(1+2+3+4+5+6+7+8+9+10)²=55²=3025 $1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} + 6^{3} + 7^{3} + 8^{3} + 9^{3} + 10^{3} = {(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)}^{2} = 55^{2} = 3025$ 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'

Answer 4: Five natural numbers, which are multiples of 3, are 3, 6, 9, 12 and 15.
Cubes of these five numbers are:

33 = 3×3×3 = 2763 = 6×6×6 = 21693 = 9×9×9 = 729123 = 12×12×12 = 1728153 = 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 × 125It 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'.

Answer 5: 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:
$64 = 3 \times 21 + 1$ , which is of the form (3n + 1) for n = 21
$343 = 3 \times 114 + 1$ , which is of the form (3n + 1) for n = 114
$1000 = 3 \times 333 + 1,$ which is of the form (3n + 1) for n = 333
$2197 = 3 \times 732 + 1,$ which is of the form (3n + 1) for n = 732
$4096 = 3 \times 1365 + 1,$ 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'.

Answer 6: 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:
$125 = 3 \times 41 + 2$ , which is of the form (3n + 2) for n = 41
$512 = 3 \times 170 + 2$ , which is of the form (3n + 2) for n = 170
$1331 = 3 \times 443 + 2,$ which is of the form (3n + 2) for n = 443
$2744 = 3 \times 914 + 2,$ which is of the form (3n + 2) for n = 914
$4913 = 3 \times 1637 + 2,$ 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 7³'.

Answer 7: 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 7³. Proceed as follows:
$343 = 7^{3} \times 1$
2744=14³=14×14×14=(7×2)×(7×2)×(7×2)=7×7×7)×(2×2×2)=7³×2³
9261=21³=21×21×21=(7×3)×(7×3)×(7×3)=(7×7×7)×(3×3×3)=7³×3³
21952=28³=28×28×28=(7×4)×(7×4)×(7×4)= (7×7×7)×(4×4×4)=7³×4³
42875=35³=35×35×35=(7×5)×(7×5)×(7×5)=(7×7×7)×(5×5×5)=7³×5³
Hence, the cube of multiple of 7 is a multiple of 7³.

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

Answer 8: (i) On factorising 64 into prime factors, we get
$64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2$
Group the factors in triples of equal factors as:
$64 = \{2 \times 2 \times 2\} \times \{2 \times 2 \times 2\}$
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:
$216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3$
Group the factors in triples of equal factors as:
$216 = \{2 \times 2 \times 2\} \times \{3 \times 3 \times 3\}$
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:
$243 = 3 \times 3 \times 3 \times 3 \times 3$
Group the factors in triples of equal factors as:
$243 = \{3 \times 3 \times 3\} \times 3 \times 3$
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:
$1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5$
Group the factors in triples of equal factors as:
$1000 = \{2 \times 2 \times 2\} \times \{5 \times 5 \times 5\}$
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:
$1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$
Group the factors in triples of equal factors as:
$1728 = \{2 \times 2 \times 2\} \times \{2 \times 2 \times 2\} \times \{3 \times 3 \times 3\}$
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:
$3087 = 3 \times 3 \times 7 \times 7 \times 7$
Group the factors in triples of equal factors as:
$3087 = 3 \times 3 \times \{7 \times 7 \times 7\}$
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:
$4608 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3$
Group the factors in triples of equal factors as:
$4608 = \{2 \times 2 \times 2\} \times \{2 \times 2 \times 2\} \times \{2 \times 2 \times 2\} \times 3 \times 3$
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:
$106480 = 2 \times 2 \times 2 \times 2 \times 5 \times 11 \times 11 \times 11$
Group the factors in triples of equal factors as:
$106480 = \{2 \times 2 \times 2\} \times 2 \times 5 \times \{11 \times 11 \times 11\}$
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:
$166375 = 5 \times 5 \times 5 \times 11 \times 11 \times 11$
Group the factors in triples of equal factors as:
$166375 = \{5 \times 5 \times 5\} \times \{11 \times 11 \times 11\}$
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:
$456533 = 7 \times 7 \times 7 \times 11 \times 11 \times 11$
Group the factors in triples of equal factors as:
$456533 = \{7 \times 7 \times 7\} \times \{11 \times 11 \times 11\}$
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

Answer 9: 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

Answer 10: 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 (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). 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

Answer 11: (i) On factorising 675 into prime factors, we get:
$675 = 3 \times 3 \times 3 \times 5 \times 5$
On grouping the factors in triples of equal factors, we get:
$675 = \{3 \times 3 \times 3\} \times 5 \times 5$
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:
$1323 = 3 \times 3 \times 3 \times 7 \times 7$
On grouping the factors in triples of equal factors, we get:
$675 = \{3 \times 3 \times 3\} \times 5 \times 5$
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:
$2560 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5$
On grouping the factors in triples of equal factors, we get:
$2560 = \{2 \times 2 \times 2\} \times \{2 \times 2 \times 2\} \times \{2 \times 2 \times 2\} \times 5$
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 $5 \times 5 = 25$ , 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:
$7803 = 3 \times 3 \times 3 \times 17 \times 17$
On grouping the factors in triples of equal factors, we get:
$7803 = \{3 \times 3 \times 3\} \times 17 \times 17$
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:
$107811 = 3 \times 3 \times 3 \times 3 \times 11 \times 11 \times 11$
On grouping the factors in triples of equal factors, we get:
$107811 = \{3 \times 3 \times 3\} \times 3 \times \{11 \times 11 \times 11\}$
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 $3 \times 3 = 9$ , 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:
$35721 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 7 \times 7$
On grouping the factors in triples of equal factors, we get:
$35721 = \{3 \times 3 \times 3\} \times \{3 \times 3 \times 3\} \times 7 \times 7$
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

Answer 12: (i) On factorising 675 into prime factors, we get:
$675 = 3 \times 3 \times 3 \times 5 \times 5$
On grouping the factors in triples of equal factors, we get:
$675 = \{3 \times 3 \times 3\} \times 5 \times 5$
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 $5 \times 5 = 25$ , 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:
$8640 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5$
On grouping the factors in triples of equal factors, we get:
$8640 = \{2 \times 2 \times 2\} \times \{2 \times 2 \times 2\} \times \{3 \times 3 \times 3\} \times 5$
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:
$1600 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5$
On grouping the factors in triples of equal factors, we get:
$1600 = \{2 \times 2 \times 2\} \times \{2 \times 2 \times 2\} \times 5 \times 5$
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 ( $5 \times 5 = 25$ ), 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:
$8788 = 2 \times 2 \times 13 \times 13 \times 13$
On grouping the factors in triples of equal factors, we get:
$8788 = 2 \times 2 \times \{13 \times 13 \times 13\}$
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 ( $2 \times 2 = 4$ ), 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:
$7803 = 3 \times 3 \times 3 \times 17 \times 17$
On grouping the factors in triples of equal factors, we get:
$7803 = \{3 \times 3 \times 3\} \times 17 \times 17$
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 ( $17 \times 17 = 289$ ), 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:
$107811 = 3 \times 3 \times 3 \times 3 \times 11 \times 11 \times 11$
On group the factors in triples of equal factors, we get:
$107811 = \{3 \times 3 \times 3\} \times 3 \times \{11 \times 11 \times 11\}$
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:
$35721 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 7 \times 7$
On grouping the factors in triples of equal factors, we get:
$35721 = \{3 \times 3 \times 3\} \times \{3 \times 3 \times 3\} \times 7 \times 7$
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 ( $7 \times 7 = 49$ ), 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:
$243000 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5$
On grouping the factors in triples of equal factors, we get:
$243000 = \{2 \times 2 \times 2\} \times \{3 \times 3 \times 3\} \times 3 \times 3 \times \{5 \times 5 \times 5\}$
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 ( $3 \times 3 = 9$ ), 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.

Answer 13: Let us consider a number n. Then its cube would be $n^{3}$ .
If the number n is trebled, i.e., 3n, we get:
${(3 n)}^{3} = 3^{3} \times n^{3} = 27 n^{3}$
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?

Answer 14: (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:
${(3 n)}^{3} = 3^{3} \times n^{3} = 27 n^{3}$
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:
${(4 n)}^{3} = 4^{3} \times n^{3} = 64 n^{3}$
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:
${(5 n)}^{3} = 5^{3} \times n^{3} = 125 n^{3}$
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.

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FAQs on RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) - RD Sharma Solutions for Class 8 Mathematics

1. What is the formula to find the volume of a cube?

Ans. The formula to find the volume of a cube is V = (side)^3, where V represents the volume and side represents the length of one side of the cube.

2. How can I find the cube root of a number?

Ans. To find the cube root of a number, you can use the prime factorization method. Prime factorize the number and group the prime factors in sets of three. The product of the prime factors in each set will give you the cube root of the number.

3. What is the difference between a perfect cube and a non-perfect cube?

Ans. A perfect cube is a number that can be expressed as the cube of an integer. For example, 8 is a perfect cube because it can be expressed as 2^3. On the other hand, a non-perfect cube is a number that cannot be expressed as the cube of an integer. For example, 10 is a non-perfect cube.

4. Can a cube have a decimal side length?

Ans. No, a cube cannot have a decimal side length. The side length of a cube is always a whole number. Cubes with decimal side lengths are not considered cubes, but rather cuboids.

5. How can I determine if a number is a perfect cube without calculating its cube root?

Ans. To determine if a number is a perfect cube without calculating its cube root, you can check if the number is divisible by the cubes of prime numbers. If the number is divisible by the cubes of prime numbers, then it is a perfect cube.

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RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) | RD Sharma Solutions for Class 8 Mathematics

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RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) | RD Sharma Solutions for Class 8 Mathematics

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RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) Free PDF Download

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RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) | RD Sharma Solutions for Class 8 Mathematics PDF Download

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

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.

Question 3: Observe the following pattern: 13 = 1 13 + 23 = (1 + 2)213 + 23 + 33 = (1 + 2 + 3)2

Answer 3: Extend the pattern as follows:

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'

Answer 4: Five natural numbers, which are multiples of 3, are 3, 6, 9, 12 and 15.Cubes of these five numbers are:

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'.

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'.

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 73'.

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

Question 9: Which of the following are cubes of even natural numbers?216, 512, 729, 1000, 3375, 13824

Question 10: Which of the following are cubes of odd natural numbers?125, 343, 1728, 4096, 32768, 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

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

Question 13: Prove that if a number is trebled then its cube is 27 times the cube of the given number.

Answer 13: Let us consider a number n. Then its cube would be n3n3.If the number n is trebled, i.e., 3n, we get:(3n)3=33×n3=27n33n3=33×n3=27n3It 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?

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RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) | RD Sharma Solutions for Class 8 Mathematics

Summary

Viva Questions

RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) | RD Sharma Solutions for Class 8 Mathematics

Sample Paper

shortcuts and tricks

Objective type Questions

Important questions

Previous Year Questions with Solutions

mock tests for examination

practice quizzes

video lectures

Extra Questions

RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) | RD Sharma Solutions for Class 8 Mathematics

past year papers

study material

pdf

MCQs

Semester Notes

ppt

Free

Exam

RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) Free PDF Download

Importance of RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1)

RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) Notes

RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) Class 8 Questions

Study RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) on the App

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

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.

Question 3: Observe the following pattern:
1³ = 1
1³ + 2³ = (1 + 2)²
1³ + 2³ + 3³ = (1 + 2 + 3)²

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'

Answer 4: Five natural numbers, which are multiples of 3, are 3, 6, 9, 12 and 15.
Cubes of these five numbers are:

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'.

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'.

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 7³'.

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

Question 9: Which of the following are cubes of even natural numbers?
216, 512, 729, 1000, 3375, 13824

Question 10: Which of the following are cubes of odd natural numbers?
125, 343, 1728, 4096, 32768, 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

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

Answer 13: Let us consider a number n. Then its cube would be $n^{3}$ .
If the number n is trebled, i.e., 3n, we get:
${(3 n)}^{3} = 3^{3} \times n^{3} = 27 n^{3}$
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?