The students can access the NCERT Solutions for the Exercise 7.2 of Class 8 Chapter 7, Cubes and Cube Roots, here. The PDF format which can be downloaded is also available here. The NCERT solutions for exercise 7.2 of Class 8 are prepared in such a way that going through these solutions will help students in getting rid of all the doubts related to the method of finding out the cube roots.

The NCERT textbook provides plenty of questions for the students to solve and practice. The NCERT solutions for class 8 Maths helps the students in understanding the most relevant method of answering a question. This way, the students get a clearer idea of the concept, each time they solve the questions present in the NCERT textbook.

### Access Answers to NCERT Class 8 Maths Chapter 7 – Cubes and Cube Roots Exercise 7.2 Page Number 275

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

**(i) 64**

Solution:

64 = 2×2×2×2×2×2

By grouping the factors in triplets of equal factors, 64 = (2×2×2)×(2×2×2)

Here, 64 can be grouped into triplets of equal factors,

∴ 64 = 2×2 = 4

Hence, 4 is cube root of 64.

(ii) 512

Solution:

512 = 2×2×2×2×2×2×2×2×2

By grouping the factors in triplets of equal factors, 512 = (2×2×2)×(2×2×2)×(2×2×2)

Here, 512 can be grouped into triplets of equal factors,

∴ 512 = 2×2×2 = 8

Hence, 8 is cube root of 512.

(iii) 10648

Solution:

10648 = 2×2×2×11×11×11

By grouping the factors in triplets of equal factors, 10648 = (2×2×2)×(11×11×11)

Here, 10648 can be grouped into triplets of equal factors,

∴ 10648 = 2 ×11 = 22

Hence, 22 is cube root of 10648.

### Access Answers to NCERT Class 8 Maths Chapter 7 – Cubes and Cube Roots Exercise 7.2 Page Number 275

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

**(i) 64**

Solution:

64 = 2×2×2×2×2×2

By grouping the factors in triplets of equal factors, 64 = (2×2×2)×(2×2×2)

Here, 64 can be grouped into triplets of equal factors,

∴ 64 = 2×2 = 4

Hence, 4 is cube root of 64.

(ii) 512

Solution:

512 = 2×2×2×2×2×2×2×2×2

By grouping the factors in triplets of equal factors, 512 = (2×2×2)×(2×2×2)×(2×2×2)

Here, 512 can be grouped into triplets of equal factors,

∴ 512 = 2×2×2 = 8

Hence, 8 is cube root of 512.

(iii) 10648

Solution:

10648 = 2×2×2×11×11×11

By grouping the factors in triplets of equal factors, 10648 = (2×2×2)×(11×11×11)

Here, 10648 can be grouped into triplets of equal factors,

∴ 10648 = 2 ×11 = 22

Hence, 22 is cube root of 10648.

(iv) 27000

Solution:

27000 = 2×2×2×3×3×3×3×5×5×5

By grouping the factors in triplets of equal factors, 27000 = (2×2×2)×(3×3×3)×(5×5×5)

Here, 27000 can be grouped into triplets of equal factors,

∴ 27000 = (2×3×5) = 30

Hence, 30 is cube root of 27000.

(v) 15625

Solution:

15625 = 5×5×5×5×5×5

By grouping the factors in triplets of equal factors, 15625 = (5×5×5)×(5×5×5)

Here, 15625 can be grouped into triplets of equal factors,

∴ 15625 = (5×5) = 25

Hence, 25 is cube root of 15625.

(vi) 13824

Solution:

13824 = 2×2×2×2×2×2×2×2×2×3×3×3

By grouping the factors in triplets of equal factors,

13824 = (2×2×2)×(2×2×2)×(2×2×2)×(3×3×3)

Here, 13824 can be grouped into triplets of equal factors,

∴ 13824 = (2×2× 2×3) = 24

Hence, 24 is cube root of 13824.

(vii) 110592

Solution:

110592 = 2×2×2×2×2×2×2×2×2×2×2×2×3×3×3

By grouping the factors in triplets of equal factors,

110592 = (2×2×2)×(2×2×2)×(2×2×2)×(2×2×2)×(3×3×3)

Here, 110592 can be grouped into triplets of equal factors,

∴ 110592 = (2×2×2×2 × 3) = 48

Hence, 48 is cube root of 110592.

(viii) 46656

Solution:

46656 = 2×2×2×2×2×2×3×3×3×3×3×3

By grouping the factors in triplets of equal factors,

46656 = (2×2×2)×(2×2×2)×(3×3×3)×(3×3×3)

Here, 46656 can be grouped into triplets of equal factors,

∴ 46656 = (2×2×3×3) = 36

Hence, 36 is cube root of 46656.

(ix) 175616

Solution:

175616 = 2×2×2×2×2×2×2×2×2×7×7×7

By grouping the factors in triplets of equal factors,

175616 = (2×2×2)×(2×2×2)×(2×2×2)×(7×7×7)

Here, 175616 can be grouped into triplets of equal factors,

∴ 175616 = (2×2×2×7) = 56

Hence, 56 is cube root of 175616.

(x) 91125

Solution:

91125 = 3×3×3×3×3×3×3×5×5×5

By grouping the factors in triplets of equal factors, 91125 = (3×3×3)×(3×3×3)×(5×5×5)

Here, 91125 can be grouped into triplets of equal factors,

∴ 91125 = (3×3×5) = 45

Hence, 45 is cube root of 91125.

2. State true or false.

**(i) Cube of any odd number is even.**

Solution:

False

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

Solution:

True

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

Solution:

False

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

Solution:

False

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

Solution:

False

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

Solution:

False

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

Solution:

True

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

Solution:

(i) By grouping the digits, we get 1 and 331

We know that, since, the unit digit of cube is 1, the unit digit of cube root is 1.

∴ We get 1 as unit digit of the cube root of 1331.

The cube of 1 matches with the number of second group.

∴ The ten’s digit of our cube root is taken as the unit place of smallest number.

We know that, the unit’s digit of the cube of a number having digit as unit’s place 1 is 1.

∴ ∛1331 = 11

(ii) By grouping the digits, we get 4 and 913

We know that, since, the unit digit of cube is 3, the unit digit of cube root is 7.

∴ we get 7 as unit digit of the cube root of 4913. We know 13 = 1 and 23 = 8 , 1 > 4 > 8.

Thus, 1 is taken as ten digit of cube root.

∴ ∛4913 = 17

(iii) By grouping the digits, we get 12 and 167.

We know that, since, the unit digit of cube is 7, the unit digit of cube root is 3.

∴ 3 is the unit digit of the cube root of 12167 We know 23 = 8 and 33 = 27 , 8 > 12 > 27.

Thus, 2 is taken as ten digit of cube root.

∴ ∛12167= 23

(iv) By grouping the digits, we get 32 and 768.

We know that, since, the unit digit of cube is 8, the unit digit of cube root is 2.

∴ 2 is the unit digit of the cube root of 32768. We know 33 = 27 and 43 = 64 , 27 > 32 > 64.

Thus, 3 is taken as ten digit of cube root.

∴ ∛32768= 32