Latest Updated : October 2023

If you are in search of **NCERT solutions for Class 8 Maths Chapter** **6 Cube and Cube Roots**, then you are at the right place. When it comes to **Mathematics**, it is a subject that holds value for students across various academic streams, be it science, biology, or commerce and having a very good understanding of basic math concepts will help you in achieving your goal and make it much more easier.

Here we provide detailed **NCERT solutions for Class 8 Maths** of all the chapters, **exercise wise** with **updated syllabus by NCERT**. These NCERT solutions for Class 8 Maths have been carefully crafted to provide detailed explanations, ensuring that students can grasp the concepts effectively. We also provide extra practice questions for the same, so feel free to bookmark **ncertforclass8.com** to boost your preparation.

## NCERT Solutions for Class 8 Maths Chapter **6 Cube and Cube Roots** Ex 6.1

**Class 8 Maths Chapter** **6 Cube and Cube Roots** **Exercise 6.1**

**Exercise 6.1**

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

**(i) 216**

**(ii) 128**

**(iii) 1000**

**(iv) 100**

(v) 46656

(v) 46656

Solution:

**(i) 216**

216 = 2×2×2×3×3×3

By grouping the factors in triplets of equal factors, 216 = (2×2×2)×(3×3×3)

Here, 216 can be grouped into triplets of equal factors,

∴ 216 is a perfect cube number.

**(ii) 128 **

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, and we are left with one factor – 2

∴ 128 is not a perfect cube number.

**(iii) 1000 **

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 triplets of equal factors.

∴ 1000 is a perfect cube number.

**(iv) 100**

100 = 2×2×5×5

Here, 100 cannot be grouped into triplets of equal factors.

∴ 100 is not a perfect cube number.

**(v) 46656**

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 triplets of equal factors,

∴ 46656 is a a perfect cube number.

**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**

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 triplets of equal factors.

∴ We will multiply 243 by 3 to get the perfect cube.

**(ii) 256 **

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 triplets of equal factors.

∴ We will multiply 256 by 2 to get the perfect cube.

**(iii) 72**

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 triplets of equal factors.

∴ We will multiply 72 by 3 to get the perfect cube.

**(iv) 675**

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 triplets of equal factors.

∴ We will multiply 675 by 5 to get the perfect cube

**(v) 100**

100 = 2×2×5×5

Here, 2 and 5 cannot be grouped into triplets 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 (ii) 128 (iii) 135 (iv) 192 (v) 704**

Solution:

**(i) 81**

Prime factors of 81 = 3x3x3x3

Here one factor 3 is not grouped in triplets.

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

**(ii) 128**

128 Prime factors of 128 = 2 x 2 x 2 x2×2×2×2

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 = 3 x 3 x 3 x 5

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 = 2 × 2 × 2 × 2 × 2 × 2 × 3

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 = 2 x 2 x 2 x 2×2×2×11

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?**

Solution:

Given numbers = 5 x 2 x 5

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

Therefore, the number must be multiplied by 2 x 2 x 5 = 20 to make it a perfect cube.

Hence, he needs 20 cuboids to form a cube.

**Using Class 8 NCERT solutions offers several key advantages for students:**

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