### Cubes and Cube Roots - Solutions 1

CBSE Class –VIII Mathematics

NCERT Solutions

CHAPTER - 7

Cubes and Cube Roots (Ex. 7.1)

NCERT Solutions

CHAPTER - 7

Cubes and Cube Roots (Ex. 7.1)

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

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

Ans. (i) 216

Prime factors of 216 =

Here all factors are in groups of 3’s (in triplets)

Therefore, 216 is a perfect cube number.

Here all factors are in groups of 3’s (in triplets)

Therefore, 216 is a perfect cube number.

(ii) 128

Prime factors of 128 =

Here one factor 2 does not appear in a 3’s group.

Therefore, 128 is not a perfect cube.

Here one factor 2 does not appear in a 3’s group.

Therefore, 128 is not a perfect cube.

(iii) 1000

Prime factors of 1000 = 2X2X2X5X5X5

Here all factors appear in 3’s group.

Therefore, 1000 is a perfect cube.

Here all factors appear in 3’s group.

Therefore, 1000 is a perfect cube.

(iv) 100

Prime factors of 100 = 2 x 2 x 5 x 5

Here all factors do not appear in 3’s group.

Therefore, 100 is not a perfect cube.

Here all factors do not appear in 3’s group.

Therefore, 100 is not a perfect cube.

(v) 46656

Prime factors of 46656 =

Here all factors appear in 3’s group.

Therefore, 46656 is a perfect cube.

Here all factors appear in 3’s group.

Therefore, 46656 is a perfect cube.

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

Ans. (i) 243

Prime factors of 243 =

Here 3 does not appear in 3’s group.

Therefore, 243 must be multiplied by 3 to make it a perfect cube.

Here 3 does not appear in 3’s group.

Therefore, 243 must be multiplied by 3 to make it a perfect cube.

(ii) 256

Prime factors of 256 =

Here one factor 2 is required to make a 3’s group.

Therefore, 256 must be multiplied by 2 to make it a perfect cube.

Here one factor 2 is required to make a 3’s group.

Therefore, 256 must be multiplied by 2 to make it a perfect cube.

(iii) 72

Prime factors of 72 =

Here 3 does not appear in 3’s group.

Therefore, 72 must be multiplied by 3 to make it a perfect cube.

Here 3 does not appear in 3’s group.

Therefore, 72 must be multiplied by 3 to make it a perfect cube.

(iv) 675

Prime factors of 675 =

Here factor 5 does not appear in 3’s group.

Therefore 675 must be multiplied by 5 to make it a perfect cube.

Here factor 5 does not appear in 3’s group.

Therefore 675 must be multiplied by 5 to make it a perfect cube.

(v) 100

Prime factors of 100 =

Here factor 2 and 5 both do not appear in 3’s group.

Therefore 100 must be multiplied by = 10 to make it a perfect cube.

Here factor 2 and 5 both do not appear in 3’s group.

Therefore 100 must be multiplied by = 10 to make it a perfect cube.

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

Ans. (i) 81

Prime factors of 81 =

Here one factor 3 is not grouped in triplets.

Therefore 81 must be divided by 3 to make it a perfect cube.

Here one factor 3 is not grouped in triplets.

Therefore 81 must be divided by 3 to make it a perfect cube.

(ii) 128

Prime factors of 128 = X 2

Here one factor 2 does not appear in a 3’s group.

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

Here one factor 2 does not appear in a 3’s group.

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

(iii) 135

Prime factors of 135 =

Here one factor 5 does not appear in a triplet.

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

Here one factor 5 does not appear in a triplet.

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

(iv) 192

Prime factors of 192 = 2X2X2X2X2X2X3

Here one factor 3 does not appear in a triplet.

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

Here one factor 3 does not appear in a triplet.

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

(v) 704

Prime factors of 704 = 2X2X2X2X2X2X11

Here one factor 11 does not appear in a triplet.

Therefore, 704 must be divided by 11 to make it a perfect cube.

Here one factor 11 does not appear in a triplet.

Therefore, 704 must be divided by 11 to make it a 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?

Ans. Given numbers =

Since, Factors of 5 and 2 both are not in group of three.

Therefore, the number must be multiplied by = 20 to make it a perfect cube.

Hence he needs 20 cuboids.

Therefore, the number must be multiplied by = 20 to make it a perfect cube.

Hence he needs 20 cuboids.