NCERT Solutions Class 8th Maths Chapter – 6 Cubes and Cube Roots Exercise – 6.1

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

(i) 216

Solution: 

By resolving 216 into prime factor,
216 = 2×2×2×3×3×3
By grouping the factors in triplets of eq Hence, 216 is a cube of 6.

(ii) 128

Solution:

By resolving 128 into prime factor,
128 = 2×2×2×2×2×2×2
By grouping the factors in triplets of equal factors, 128 = (2×2×2)×(2×2×2)×2

Here, 128 cannot be grouped into triplets of equal factors, we are left of with one factors 2 .
∴ 128 is not a perfect cube.

(iii) 1000

Solution: 

By resolving 1000 into prime factor,

1000 = 2×2×2×5×5×5
By grouping the factors in triplets of equal factors, 1000 = (2×2×2)×(5×5×5)

Here, 1000 can be grouped into triples of equal factors,
∴ 1000 = (2×5) = 10
Hence, 1000 is a cube of 10.

(iv) 100

Solution: 

By resolving 100 into prime factor,

100 = 2×2×5×5

Here, 100 cannot be grouped into triplets of equal factors.
∴ 100 is not a perfect cube.

(v) 46656

Solution: 

By resolving 46656 into prime factor,

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 triples of equal factors,
∴ 46656 = (2×2×3×3) = 36
Hence, 46656 is a cube of 36.

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

(i) 243

Solution: 

By resolving 243 into prime factor,

243 = 3×3×3×3×3
By grouping the factors in triplets of equal factors, 243 = (3×3×3)×3×3

Here, 3 cannot be grouped into triples of equal factors.
∴ We will multiply 243 by 3 to get the perfect cube.

(ii) 256

Solution: 

By resolving 256 into prime factor,

256 = 2×2×2×2×2×2×2×2
By grouping the factors in triplets of equal factors, 256 = (2×2×2)×(2×2×2)×2×2

Here, 2 cannot be grouped into triples of equal factors.
∴ We will multiply 256 by 2 to get the perfect cube.

(iii) 72

Solution: 

By resolving 72 into prime factor,

72 = 2×2×2×3×3
By grouping the factors in triplets of equal factors, 72 = (2×2×2)×3×3

Here, 3 cannot be grouped into triples of equal factors.
∴ We will multiply 72 by 3 to get the perfect cube.

(iv) 675

Solution:

By resolving 675 into prime factor,

675 = 3×3×3×5×5
By grouping the factors in triplets of equal factors, 675 = (3×3×3)×5×5

Here, 5 cannot be grouped into triples of equal factors.
∴ We will multiply 675 by 5 to get the perfect cube.

(v) 100

Solution: By resolving 100 into prime factor,

100 = 2×2×5×5

Here, 2 and 5 cannot be grouped into triples of equal factors.
∴ We will multiply 100 by (2×5) 10 to get the perfect cube.

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

(i) 81

Solution: 

By resolving 81 into prime factor,

81 = 3×3×3×3
By grouping the factors in triplets of equal factors, 81 = (3×3×3)×3

Here, 3 cannot be grouped into triples of equal factors.
∴ We will divide 81 by 3 to get the perfect cube.

(ii) 128

Solution: 

By resolving 128 into prime factor,

128 = 2×2×2×2×2×2×2
By grouping the factors in triplets of equal factors, 128 = (2×2×2)×(2×2×2)×2

Here, 2 cannot be grouped into triples of equal factors.
∴ We will divide 128 by 2 to get the perfect cube.

(iii) 135

Solution:

By resolving 135 into prime factor,

135 = 3×3×3×5
By grouping the factors in triplets of equal factors, 135 = (3×3×3)×5

Here, 5 cannot be grouped into triples of equal factors.
∴ We will divide 135 by 5 to get perfect cube.

(iv) 192

Solution: 

By resolving 192 into prime factor,

192 = 2×2×2×2×2×2×3
By grouping the factors in triplets of equal factors, 192 = (2×2×2)×(2×2×2)×3

Here, 3 cannot be grouped into triplets of equal factors.
∴ We will divide 192 by 3 to get perfect cube.

(v) 704

Solution: 

By resolving 704 into prime factor,

704 = 2×2×2×2×2×2×11
By grouping the factors in triplets of equal factors, 704 = (2×2×2)×(2×2×2)×11

Here, 11 cannot be grouped into triples of equal factors.
∴ We will divide 704 by 11 to get the perfect cube.

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, side of cube is 5 cm, 2 cm and 5 cm.
∴ Volume of cube = 5×2×5 = 50

50 = 2×5×5

Here, 2 , 5 and 5 cannot be grouped into triplets of equal factors.∴ We will multiply 50 by (2×2×5) 20 to get perfect cube. Hence, 20 cuboid is needed.