These NCERT Solutions for Class 8 Maths Chapter 7 Cube and Cube Roots Ex 7.2 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 8 Maths Chapter 7 Cube and Cube Roots Exercise 7.2

Question 1.

Find the cube root of each of the following numbers by prime factorisation method.

(i) 64

(ii) 512

(iii) 10648

(iv) 27000

(v) 15625

(vi) 13824

(vii) 110592

(viii) 46656

(ix) 175616

(x) 91125

Solution:

(i) 64

On grouping the factors in triplets, we get

64 = 2^{3} × 2^{3}

64 = (2 × 2)^{3} = 4^{3}

\(\sqrt[3]{64}\) = 4

(ii) 512

On grouping the factors in triplets, we get

512 = 2^{3} × 2^{3} × 2^{3}

= (2 × 2 × 2)^{3}

= 8^{3}

\(\sqrt[3]{512}\) = 8

(iii) 10648

On grouping the factors in triplets, we get

10648 = 2^{3} × 11^{3}

= (2 × 11)^{3}

= 22^{3}

\(\sqrt[3]{10648}\) = 22

(iv) 27000

On grouping the factors in triplets, we get

27000 = 2^{3} × 3^{3} × 5^{3}

= (2 × 3 × 5)^{3}

= 30^{3}

\(\sqrt[3]{27000}\) = 30

(v) 15625

On grouping the factors in triplets, we get

15625 = 5^{3} × 5^{3}

= (5 × 5)^{3}

= 25^{3}

\(\sqrt[3]{15625}\) = 25

(vi) 13824

On grouping the factors in triplets, we get

13824 = 2^{3} × 2^{3} × 2^{3} × 3^{3}

= (2 × 2 × 2 × 3)^{3}

= 24^{3}

\(\sqrt[3]{13824}\) = 24

(vii) 110592

On grouping the factors in triplets, we get

110592 = 2^{3} × 2^{3} × 2^{3} × 2^{3} × 3^{3}

= (2 × 2 × 2 × 2 × 3)^{3}

= 48^{3}

\(\sqrt[3]{110592}\) = 48

(viii) 46656

On grouping the factors in triplets, we get

46656 = 2^{3} × 2^{3} × 3^{3} × 3^{3}

= (2 × 2 × 3 × 3)^{3}

= 36^{3}

\(\sqrt[3]{46656}\) = 36

(ix) 175616

On grouping the factors in triplets, we get

175616 = 2^{3} × 2^{3} × 2^{3} × 7^{3}

= (2 × 2 × 2 × 7)^{3}

= 56^{3}

\(\sqrt[3]{175616}\) = 56

(x) 91125

On grouping the factors in triplets, we get

91125 = 3^{3} × 3^{3} × 5^{3}

= (3 × 3 × 5)^{3}

= 45^{3}

\(\sqrt[3]{91125}\) = 45

Question 2.

State true or false.

(i) Cube of any odd number is even.

(ii) A perfect cube does not end with two zeros.

(iii) If the square of a number ends with 5, then its cube ends with 25.

(iv) There is no perfect cube which ends with 8.

(v) The cube of a two-digit number may be a three-digit number.

(vi) The cube of a two-digit number may have seven or more digits.

(vii) The cube of a single-digit number may be a single-digit number.

Solution:

(i) False

(ii) True

(iii) False [15^{2} = 225; 15^{3} = 3375]

(iv) False [12^{3} = 1728]

(v) False [10^{3} = 1000]

(vi) False [99^{3} = 970299]

(vii) True [2^{3} = 8]

Question 3.

You are told that 1,331 is a perfect cube. Can you guess without factorization what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.

Solution:

(i) Separating the given number 1331 into two groups.

1 and 331

331 ends in 1

unit digit of the cube root = 1

Tens digit of the cube root = 1

\(\sqrt[3]{1331}\) = 11

(ii) Cube root of 4913

Separating the given number 4913 in two groups i.e. 4 and 913

In this case, 913 has three-digit and 4 has only one digit

The digit 3 is at its own place. We take the one’s place of the required cube root as 7.

Take the other group i.e. 4, a cube of 1 is 1, and a cube of 2 is 8. 4 lies between 1 and 8.

The smaller number among 1 and 2 is 1

The one place of 1 is 1 itself.

Take 1 as ten’s place of the cube root of 4913.

\(\sqrt[3]{4913}\) = 17

(iii) Cube root of 12167

Separating 12167 in two groups i.e. 12 and 167

The digit 7 is at its one’s place. We take the one’s place of required cube root as 3.

The unit digit of the cube root = 3

Take the other group i.e. 12 Cube of 2 is 8 and cube of 3 is 27. 12 lies between 8 and 27

The smaller among 2 and 3 is 2

The one place is 2 itself.

Take 2 as ten’s place of the cube root of 12167.

\(\sqrt[3]{12167}\) = 23

(iv) Cube root of 32768

Separating 32768 in two groups i.e. 32 and 768

Take 768

The digit 8 is at its one’s place so, the one’s place of the required cube root is 2.

Take the other group i.e. 32

The cube of 3 is 27 and the cube of 4 is 64.

32 lies between 27 and 64.

The smaller number between 3 and 4 is 3

Take 3 as ten’s place of the cube root of 32768

\(\sqrt[3]{32768}\) = 32