**Download NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.1 – Cube and Cube roots**

**NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.1 – Cube and Cube roots, has been designed by the NCERT to test the knowledge of the student on the following topics :**

- Introduction
- Cubes

Some interesting patterns

1. Adding consecutive odd numbers

2. Cubes and their prime factors

Smallest multiple that is a perfect cube

**Q.1 Which of the following numbers are not perfect cubes?**

**(i) 216
(ii) 128
(iii) 1000
(iv) 100
(v) 46656**

**Solution: **

**(i)** Prime factorisation of 216:

216 = __2 × 2 × 2__ × __3 × 3 × 3 __= 2^{3} × 3^{3
}Here, factors of 216 can be grouped in triples.

Therefore, 216 is a perfect cube.

**(ii)** Prime factorisation of 128:

128 = __2 × 2 × 2__ × __2 × 2 × 2__ × 2

Here, all factors of 128 cannot be grouped in triples as one 2 is remaining.

Therefore, 128 is not a perfect cube.

**(iii)** Prime factorisation of 1000:

1000 = __2 × 2 × 2 __× __5 × 5 × 5
__Here, all factors of 1000 can be grouped in triples

Therefore, 1000 is a perfect cube.

**(iv)** Prime factorisation of 100:

100 = 2 × 2 × 5 × 5

Here, factors of 100 cannot be grouped in triples

Therefore, 100 is not a perfect cube.

**(v)** Prime factorisation of 46656:

46656 = __2 × 2 × 2__ × __2 × 2 × 2__ × __3 × 3 × 3__ × __3 × 3 × 3
__Here, factors of 46656 can be grouped in triples.

Therefore, 46656 is a perfect cube.

** **

**Q.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 = __3 × 3 × 3__ × 3 × 3

Here, after grouping the factors into triples, two 3’s is left. To make them a group of triples one 3 has to be multiplied

243 × 3 = __3 × 3 × 3__ × __3 × 3 × 3__ = 729 is a perfect cube.

Hence, the smallest natural number by which 243 must be multiplied to make it a perfect cube is 3.

**(ii)** 256 = __2 × 2 × 2__ × __2 × 2 × 2__ × 2 × 2

Here, after grouping the factors into triples, two 2’s is left. To make them a group of triples one 2 has to be multiplied

256 × 2 = __2 × 2 × 2__ × __2 × 2 × 2__ × __2 × 2 × 2__ = 512 is a perfect cube.

Hence, the smallest natural number by which 256 must be multiplied to make it a perfect cube is 2.

**(iii)** 72 = __2 × 2 × 2__ × 3 × 3

Here, after grouping the factors into triples, two 3’s is left. To make them a group of triples one 3 has to be multiplied

72 × 3 = __2 × 2 × 2__ × __3 × 3 × 3__ = 216 is a perfect cube.

Hence, the smallest natural number by which 72 must be multiplied to make it a perfect cube is 3.

**(iv)** 675 = __3 × 3 × 3__ × 5 × 5

Here, after grouping the factors into triples, two 5’s is left. To make them a group of triples one 5 has to be multiplied

675 × 5 = __3 × 3 × 3__ × __5 × 5 × 5__ = 3375 is a perfect cube.

Hence, the smallest natural number by which 675 must be multiplied to make it a perfect cube is 5.

**(v)** 100 = 2 × 2 × 5 × 5

Here, factors of 100 cannot be grouped in triples. To make them a group of triples one 2 and one 5 has to be multiplied

100 × 2 × 5 = __2 × 2 × 2__ × __5 × 5 × 5__ = 1000 is a perfect cube

Hence, the smallest natural number by which 100 must be multiplied to make it a perfect cube is 2 × 5 = 10.

**Q.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 = __3 × 3 × 3__ × 3

Here, one 3 is left after grouping factors in triples.

81 has to be divided by 3, to make it a perfect cube.

81 ÷ 3 = 27 = __3 × 3 × 3__ is a perfect cube.

Hence, the smallest number by which 81 must be divided to make it a perfect cube is 3

**(ii)** 128 = __2 × 2 × 2__ × __2 × 2 × 2__ × 2

Here, one 2 is left after grouping factors in triples

128 has to be divided by 2, to make it a perfect cube.

128 ÷ 2 = 64 = __2 × 2 × 2__ × __2 × 2 × 2__ is a perfect cube.

Hence, the smallest number by which 128 must be divided to make it a perfect cube is 2

**(iii)** 135 = __3 × 3 × 3__ × 5

Here, one 5 is left after grouping factors in triples

135 has to be divided by 5, to make it a perfect cube.

135 ÷ 5 = 27 = __3 × 3 × 3__ is a perfect cube.

Hence, the smallest number by which 135 must be divided to make it a perfect cube is 5

**(iv)** 192 = __2 × 2 × 2__ × __2 × 2 × 2__ × 3

Here, one 3 is left after grouping the factors in triples

192 has to be divided by 3, to make it a perfect cube.

192 ÷ 3 = 64 = __2 × 2 × 2__ × __2 × 2 × 2__ is a perfect cube.

Hence, the smallest number by which 192 must be divided to make it a perfect cube is 3

**(v)** 704 = __2 × 2 × 2__ × __2 × 2 × 2__ × 11

Here, one 11 is left after grouping the factors in triples

704 has to be divided by 11, to make it a perfect cube

704 ÷ 11 = 64 = __2 × 2 × 2__ × __2 × 2 × 2__ is a perfect cube.

Hence, the smallest number by which 704 must be divided to make it a perfect cube is 11

** **

**Q.4 Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?**

**Solution:**

Given, Parikshit makes a cuboid of sides 5 cm, 2 cm, 5 cm

To form a cube with these cuboids:

LCM of 5, 2 and 5 = 10

Therefore, length of each side of cube should be 10 cm

For this arrangement, we have to put, 2 cuboids along with its length ( 2×5=10 ),

5 along with its width ( 5×2=10 ), and 2 along with its height ( 2×5=10 )

Therefore, total cuboids required = 2 × 5 × 2 = 20

**The next Exercise for** **NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.2 – Cube and Cube roots can be accessed by clicking here**

**Download NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.1 – Cube and Cube roots**

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