These NCERT Solutions for Class 8 Maths Chapter 7 Cube and Cube Roots Ex 7.1 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 8 Maths Chapter 7 Cube and Cube Roots Exercise 7.1

Question 1.

Which of the following numbers are not perfect cubes?

(i) 216

(ii) 128

(iii) 1000

(iv) 100

(v) 46656

Solution:

(i) 216

216 = 2^{3} × 3^{3}

Each prime factors appear in triplets, so 216 is a perfect cube.

(ii) 128

128 = 2^{3} × 2^{3} × 2

2 remains after grouping in triplets, so 128 is not a perfect cube.

(iii) 1000

1000 = 2^{3} × 5^{3}

Each prime factor appears in triplets so 1000 is a perfect cube.

(iv) 100

100 = 2^{2} × 5^{2}

As we do not get any triplets, So 100 is not a perfect cube.

(v) 46656

46656 = 2^{3} × 2^{3} × 3^{3} × 3^{3}

Each prime factor appears in triplets, so 46656 is a perfect cube.

Question 2.

Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.

(i) 243

(ii) 256

(iii) 72

(iv) 675

(v) 100

Solution:

(i) 243

243 = 3^{3} × 3^{2}

On grouping the factors in triplets, the prime factor 3 does not appear in a group of three.

To make it a perfect cube, we need one more 3.

The number 243 should be multiplied by 3 to get a perfect cube.

(ii) 256

256 = 2^{3} × 2^{3} × 2^{2}

On grouping the factors in triplets, the prime factor 2 does not appears in a group of three.

To make it a perfect cube, we need one more 2.

The number 256 should be multiplied by 2 to get a perfect cube.

(iii) 72

72 = 2^{3} × 3^{2}

On grouping, the factors in triplets the prime factor 3 does not appear in a group of 3.

To make it a perfect cube, we multiply it by 3.

The number 72 should be multiplied by 3 to get a perfect cube.

(iv) 675

675 = 3^{3} × 5^{2}

On grouping the factors in triplets, the prime factor 5 does not appear in a group of three.

To make it a perfect cube, we multiply it by 5.

The number 675 should be multiplied by 5 to get a perfect cube.

(v) 100

100 = 2^{2} × 5^{2}

On grouping the factors in triplets, the prime factors 2 and 5 do not appear in a group of three.

To make it a perfect cube we need one 2 and one 5 (2 × 5 = 10)

The number 100 should be multiplied by 10 to get a perfect cube.

Question 3.

Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.

(i) 81

(ii) 128

(iii) 135

(iv) 192

(v) 704

Solution:

(i) 81

81 = 3^{3} × 3

On grouping the factors in triplets, we find that three is no triplet of 3.

To make it a perfect cube, we divide it by 3.

(ii) 128

128 = 2^{3} × 2^{3} × 2

On grouping the factors in triplets, we find that there is no triplet of 2.

To make it a perfect cube, we divide it by 2.

(iii) 135

135 = 3^{3} × 5

On grouping the factors in triplets, the prime factor 5 does not appear in a group of three.

To make it a perfect cube, 135 should be divided by 5.

(iv) 192

192 = 2^{3} × 2^{3} × 3

On grouping the factors in triplets, the prime factor 3 does not appear in a group of three.

To make it a perfect cube, 192 should be divided by 3.

(v) 704

704 = 2^{3} × 2^{3} × 11

On grouping the factors in triplets, the prime factor 11 does not appear in a group of three.

To make it a perfect cube, 704 should be divided by 11.

Question 4.

Parikshit makes a cuboid of plasticine on sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?

Solution:

Sides of the cuboid are 5 cm, 2 cm, and 5 cm

Volume of the cuboid = 5 cm × 2 cm × 5 cm

To form it as a cube, its dimension should be in the group of triplets.

Since there is only one 2 and only two 5 in the prime factorization.

We need 2 × 2 × 5 = 20 such cuboids to make a cube.