Open App

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.

The document RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) | RD Sharma Solutions for Class 8 Mathematics is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.

All you need of Class 8 at this link: Class 8

	RD Sharma Solutions for Class 8 Mathematics 88 docs

RD Sharma Solutions for Class 8 Mathematics

88 docs

Join Course for Free

Top Courses for Class 8

View all

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.

Related Exams

Class 8

About this Document

	4.61/5 Rating
	Nov 30, 2024 Last updated

Document Description: RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) for Class 8 2024 is part of RD Sharma Solutions for Class 8 Mathematics preparation. The notes and questions for RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) have been prepared according to the Class 8 exam syllabus. Information about RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) covers topics like and RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) Example, for Class 8 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1).

Introduction of RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) in English is available as part of our RD Sharma Solutions for Class 8 Mathematics for Class 8 & RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) in Hindi for RD Sharma Solutions for Class 8 Mathematics course. Download more important topics related with notes, lectures and mock test series for Class 8 Exam by signing up for free. Class 8: RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) | RD Sharma Solutions for Class 8 Mathematics

Description

Full syllabus notes, lecture & questions for RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) | RD Sharma Solutions for Class 8 Mathematics - Class 8 | Plus excerises question with solution to help you revise complete syllabus for RD Sharma Solutions for Class 8 Mathematics | Best notes, free PDF download

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

In this doc you can find the meaning of RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) defined & explained in the simplest way possible. Besides explaining types of RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) tests, examples and also practice Class 8 tests

	RD Sharma Solutions for Class 8 Mathematics 88 docs

RD Sharma Solutions for Class 8 Mathematics

88 docs

Join Course for Free

Download as PDF

Explore Courses for Class 8 exam

Top Courses for Class 8

Explore Courses

Signup for Free!

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.

Start learning for Free

10M+ students study on EduRev

past year papers

Exam

study material

video lectures

Previous Year Questions with Solutions

Objective type Questions

Free

Semester Notes

shortcuts and tricks

mock tests for examination

Viva Questions

MCQs

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

practice quizzes

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

Extra Questions

Sample Paper

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

Summary

ppt

pdf

Important questions

;

Additional Information about RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) for Class 8 Preparation

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

The RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) is an invaluable resource that delves deep into the core of the Class 8 exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) now and kickstart your journey towards success in the Class 8 exam.

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

The importance of RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) cannot be overstated, especially for Class 8 aspirants. This document holds the key to success in the Class 8 exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

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) Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1). It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) Notes on EduRev are your ultimate resource for success.

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

The "RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) Class 8 Questions" guide is a valuable resource for all aspiring students preparing for the Class 8 exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

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

Students of Class 8 can study RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1), students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) is prepared as per the latest Class 8 syllabus.

Education Revolution

Signup to see your scores go up within 7 days!

Access 1000+ FREE Docs, Videos and Tests

Continue with Google

Takes less than 10 seconds to signup

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?

RD Sharma Solutions for Class 8 Mathematics

Top Courses for Class 8

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

RD Sharma Solutions for Class 8 Mathematics

Top Courses for Class 8

past year papers

Exam

study material

video lectures

Previous Year Questions with Solutions

Objective type Questions

Free

Semester Notes

shortcuts and tricks

mock tests for examination

Viva Questions

MCQs

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

practice quizzes

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

Extra Questions

Sample Paper

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

Summary

ppt

pdf

Important questions

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?