  # NCERT Solutions for Class 8 Maths Chapter 7 (Ex 7.1) Cubes and Cube Roots

## NCERT Solutions for Class 8 Chapter 7 Cubes and Cube Roots -Free PDF Download

Free PDF download of NCERT Solutions Maths Class 8 Solutions Chapter 7 – Cubes and Cube Roots solved by Expert Maths Teachers on CoolGyan.Org. All Chapter 7 – Cubes and Cube Roots Questions with Solutions for NCERT to help you to revise complete Syllabus and Score More marks.
Maths Revision Notes for Class 8

 Chapter Name Cubes and Cube Roots Chapter Chapter 7 Exercise Exercise 7.1 Class Class 8 Subject Maths NCERT Solutions Board CBSE TEXTBOOK CBSE NCERT Category NCERT Solutions

# NCERT SOLVED

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.

(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.

(iii) 1000 Prime factors of 1000 = 2X2X2X5X5X5
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.

(v) 46656 Prime factors of 46656 = 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.

(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.

(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.

(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.
(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.

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.

(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.

(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.

(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.
(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.

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.