# 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
2. Cubes and their prime factors
Smallest multiple that is a perfect cube

### 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

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 = 23 × 33
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

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