NCERT Solutions for Class 8 Maths Chapter 6 Cubes and Cube Roots

• Exercise 6.1: Exercise 6.1 Focuses on understanding the concept of Cubes. Problems related to finding the Cubes of given numbers and Identifying perfect Cubes and solving simple Cube-related problems.

• Exercise 6.2: Exercise 6.2 Deals with the concept of Cube Roots  and finding the Cube Roots  of given numbers using prime factorization.

Access NCERT Solutions for Class 8 Maths Chapter 6  – Cubes and Cube Roots

Exercise 6.1

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

(a) 216

Ans: Prime factorisation of 216 is 

$\therefore 216=2\times 2\text{ }\times 2\times 3\times 3\times 3=\text{ }{{2}^{3}}\times {{3}^{3}}$

Here, as each prime factor 2 and 3 are appearing as many times as a perfect triplet, 216 is a perfect cube.

(b) 128

Ans: Prime factorisation of 128 is

$ 128=2\times 2\times 2\times 2\times 2\times 2\times 2={{2}^{3}}\times {{2}^{3}}\times 2$ 

Here, the prime factor $2$ is appearing in two triplets and an extra $2$. Thus, $128$ is not a perfect cube. 

(c) 1000

Ans: The prime factorization of 1000 is

$ 1000=2\times 2\times 2\times 5\times 5\times 5={{2}^{2}}\times {{5}^{2}}$ 

Here, each prime factor is appearing as a perfect triplet, thus, 1000 is a perfect cube.

(d) 100

Ans: The prime factorisation of 100 is as follows.

 $ 100=2\times 2\times 5\times 5$ 

Here, each prime factor is not appearing as a perfect triplet. Thus, 100 is not a perfect cube.

(e) 46656

Ans: Prime factorisation of 46656 is 

 $ 46656\text{ }=\text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }3\text{ }\times \text{ }3\text{ }\times \text{ }3\text{ }\times \text{ }3\text{ }\times \text{ }3\text{ }\times \text{ }3$ 

Here, as each prime factor is appearing as a perfect triplet, thus, 46656 is a perfect cube. 

The numbers whose factors are not in the triplet are not perfect cubes

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

(a) 243 

Ans: The prime factorisation of $243$ is  $ 243=\text{ }3\text{ }\times \text{ }3\text{ }\times \text{ }3\text{ }\times \text{ }3\text{ }\times \text{ }3$  Here, two 3s are extra which are not in a triplet. To make 243 a cube, one more 3 is required. 

In that case,  $ 243\times 3=3\times 3\times 3\times 3\times 3\times 3=729$  is a perfect cube. 

Therefore, the smallest natural number by which 243 should be multiplied to make it a perfect cube is 3.

(b) 256 

Ans: The prime factorisation of  $ ~256=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2$ .

Here, two 2s are extra which are not in a triplet. To make 256 a cube, one more 2 is required. Then, we obtain 

$ 256\text{ }\times \text{ }2\text{ }=\text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }=\text{ }512$ , which is a perfect cube.

Therefore, the smallest natural number by which 256 should be multiplied to make it a perfect cube is 2. 

(c) 72 

Ans: $ 72\text{ }=\text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }3\text{ }\times \text{ }3$ 

Here, two 3s are extra which are not in a triplet. To make 72 a perfect cube, one more 3 is required. 

Thus, we obtain  $ 72\text{ }\times \text{ }3\text{ }=\text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }3\text{ }\times \text{ }3\text{ }\times \text{ }3\text{ }=\text{ }216$  which is a perfect cube. 

Therefore, the smallest natural number by which 72 should be multiplied to make it a perfect cube is 3. 

(d) 675

Ans: $ 675\text{ }=\text{ }3\text{ }\times \text{ }3\text{ }\times \text{ }3\text{ }\times \text{ }5\text{ }\times \text{ }5$  Here, two 5s are extra which are not in a triplet. To make 675 a perfect cube, one more 5 is required. 

Then, we obtain  $ 675\text{ }\times \text{ }5\text{ }=\text{ }3\text{ }\times \text{ }3\text{ }\times \text{ }3\text{ }\times \text{ }5\text{ }\times \text{ }5\text{ }\times \text{ }5\text{ }=\text{ }3375$  which is a perfect cube. 

Therefore, the smallest natural number by which 675 should be multiplied to make it a perfect cube is 5. 

(e) 100 

Ans: $ 100\text{ }=\text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }5\text{ }\times \text{ }5$ . Here, two 2s and two 5s are extra which are not in a triplet. To make 100 a cube, we require one more 2 and one more 5. Then, we obtain  $ 100\text{ }\times \text{ }2\text{ }\times \text{ }5\text{ }=\text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }5\text{ }\times \text{ }5\text{ }\times \text{ }5=1000$ which is a perfect cube.

Therefore, the smallest natural number by which 100 should be multiplied to make it a perfect cube is  $ 2\text{ }\times \text{ }5\text{ }=\text{ }10$ . 

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

(a) 81

Ans: $ 81\text{ }=\text{ }\underline{3\text{ }\times \text{ }3\text{ }\times \text{ }3}\text{ }\times \text{ }3$ . Here, one 3 is extra which is not in a triplet. Dividing 81 by 3, will make it a perfect cube. 

Thus,  $ 81\text{ }\div \text{ }3\text{ }=\text{ }27\text{ }=\text{ }3\text{ }\times \text{ }3\text{ }\times \text{ }3$  is a perfect cube.

Hence, the smallest number by which 81 should be divided to make it a perfect cube is 3. 

(b) 128

Ans: $ 128\text{ }=\text{ }\underline{2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }}\times \text{ }\underline{2\text{ }\times \text{ }2\text{ }\times \text{ }2}\text{ }\times \text{ }2$ . Here, one 2 is extra which is not in a triplet. If we divide 128 by 2, then it will become a perfect cube. Thus,  $ 128\text{ }\div \text{ }2\text{ }=\text{ }64\text{ }=\text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2$  is a perfect cube.

Hence, the smallest number by which 128 should be divided to make it a perfect cube is 2. 

(c) 135 

Ans: $ 135\text{ }=\text{ }\underline{3\text{ }\times \text{ }3\text{ }\times \text{ }3}\text{ }\times \text{ }5$ . Here, one 5 is extra which is not in a triplet. If we divide 135 by 5, then it will become a perfect cube. 

Therefore,  $ 135\text{ }\div \text{ }5\text{ }=\text{ }27\text{ }=\text{ }3\text{ }\times \text{ }3\text{ }\times \text{ }3$  is a perfect cube. 

Hence, the smallest number by which 135 should be divided to make it a perfect cube is 5. 

(d) 192

Ans:  $ 192=2\times 2\times 2\times 2\times 2\times 2\times 3$ . 

Here, one 3 is left which is not in a triplet. If we divide 192 by 3, then it will become a perfect cube. Thus,  $ 192\div 3=\text{6}4=2\times 2\times 2\times 2\times 2\times 2$  is a perfect cube. 

Therefore, the smallest number by which 192 should be divided to make it a perfect cube is 3. 

(e) 704 

Ans: $ 704=2\times 2\times 2\times 2\times 2\times 2\times 11$ . Here, one 11 is left which is not in a triplet. If we divide 704 by 11, then it will become a perfect cube. Thus,  $ 704\div 11=\text{6}4=2\times 2\times 2\times 2\times 2\times 2$  is a perfect cube. Therefore, the smallest number by which 704 should be divided to make it a perfect cube is 11. 

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: Some cuboids of size 5 × 2 × 5 are given. These cuboids when arranged to form a cube, the side of this cube is so formed that it will be a common multiple of the sides (i.e., 5, 2, and 5) of the given cuboid. 

Finding the LCM of 5, 2, and 5 we get 10. Thus, a cube of 10 cm side needs to be made. For this arrangement, we have to put 2 cuboids along with its length, 5 along with its width, and 2 along with its height. Therefore, the total cuboids required according to this arrangement = 2 × 5 × 2 = 20 With the help of 20 cuboids of such measures, the required cube is formed.

Otherwise,

Volume of the cube of sides  $ 5cm,2cm,5cm=5\text{c}m\times 2cm\times 5\text{c}m=\left( 5\times 5\times 2 \right)c{{m}^{3}}$ Here, two 5s and one 2 are extra which are not in a triplet. If we multiply this expression by  $ 2\times 2\times 5=20$ , then it will become a perfect cube. Thus, $ ~\left( 5\text{ }\times \text{ }5\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }5 \right)\text{ }=\text{ }\left( 5\text{ }\times \text{ }5\text{ }\times \text{ }5\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2 \right)\text{ }=\text{ }1000$  is a perfect cube. Hence, 20 cuboids of 5 cm, 2 cm, 5 cm are required to form a cube.

Exercise 6.2

1. Find the Cube Roots of each of the following numbers by prime factorisation method

i. $64$

Ans: Expand $64$ in factors of prime numbers.

$64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 $

$= {2^3} \times {2^3} $

Take Cube Roots on both sides of equation.

$ \because 64 = {2^3} \times {2^3} $

$\therefore \sqrt[3]{{64}} = 2 \times 2 = 4 $

The Cube Roots of $64$ is $4.$

ii. $512$

Ans: Expand $512$ in factors of prime numbers.

$512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 $

$= {2^3} \times {2^3} \times {2^3} $

Take Cube Roots on both sides of equation.

$\because 512 = {2^3} \times {2^3} \times {2^3} $

$\therefore \sqrt[3]{{512}} = 2 \times 2 \times 2 = 8 $

The Cube Roots of $512$ is $8.$

iii. $10648$

Ans: Expand $10648$ in factors of prime numbers.

$10648 = 2 \times 2 \times 2 \times 11 \times 11 \times 11 $

$= {2^3} \times {11^3} $

Take Cube Roots on both sides of equation.

$\because 10648 = {2^3} \times {11^3} $

$\therefore \sqrt[3]{{10648}} = 2 \times 11 = 22 $

The Cube Roots of $10648$ is $22.$

iv. $27000$

Ans: Expand $27000$ in factors of prime numbers.

$ 27000 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 $

$= {2^3} \times {3^3} \times {5^3} $

Take Cube Roots on both sides of equation.

$\because 27000 = {2^3} \times {3^3} \times {5^3} $

$\therefore \sqrt[3]{{27000}} = 2 \times 3 \times 5 = 30 $

The Cube Roots of $27000$ is $30.$

v. $15625$

Ans: Expand $15625$ in factors of prime numbers.

$  15625 = 5 \times 5 \times 5 \times 5 \times 5 \times 5 $

$ = {5^3} \times {5^3} $

Take Cube Roots on both sides of equation.

$\because 15625 = {5^3} \times {5^3} $

$\therefore \sqrt[3]{{15625}} = 5 \times 5 = 25 $

The Cube Roots of $15625$ is $25.$

vi. $13824$

Ans: Expand $13824$ in factors of prime numbers.

$13824 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 $

$= {2^3} \times {2^3} \times {2^3} \times {3^3} $

Take Cube Roots on both sides of equation.

$  \because 13284 = {2^3} \times {2^3} \times {2^3} \times {3^3} $

$\therefore \sqrt[3]{{13284}} = 2 \times 2 \times 2 \times 3 = 24 $

The Cube Roots of $13824$ is $24$

vii. $110592$

Ans: Expand $110592$ in factors of prime numbers.

$115092 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 $

$= {2^3} \times {2^3} \times {2^3} \times {2^3} \times {3^3} $

Take Cube Roots on both sides of equation.

$\because 110592 = {2^3} \times {2^3} \times {2^3} \times {2^3} \times {3^3} $

$\therefore \sqrt[3]{{110592}} = 2 \times 2 \times 2 \times 2 \times 3 = 48 $

The Cube Roots of $110592$ is $48.$

viii. $46656$

Ans: Expand $46656$ in factors of prime numbers.

$46656 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 $

$= {2^3} \times {2^3} \times {3^3} \times {3^3} $

Take Cube Roots on both sides of equation.

$\because 46656 = {2^3} \times {2^3} \times {3^3} \times {3^3} $

$\therefore \sqrt[3]{{46656}} = 2 \times 2 \times 3 \times 3 = 36 $

The Cube Roots of $46656$ is $36.$

ix. $175616$

Ans: Expand $175616$ in factors of prime numbers.

$175616 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7 \times 7 \times 7 $

$= {2^3} \times {2^3} \times {2^3} \times {7^3} $

Take Cube Roots on both sides of equation.

$\because 175616 = {2^3} \times {2^3} \times {2^3} \times {7^3} $

$\therefore \sqrt[3]{{175616}} = 2 \times 2 \times 2 \times 7 = 56 $

The Cube Roots of $175616$ is $56.$

x. $91125$

Ans: Expand $91125$ in factors of prime numbers.

$91125 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 $

$= {3^3} \times {3^3} \times {5^3} $

Take Cube Roots on both sides of equation.

$\because 91125 = {3^3} \times {3^3} \times {5^3} $

$\therefore \sqrt[3]{{91125}} = 3 \times 3 \times 5 = 45 $

The Cube Roots of $91125$  is $45.$

2. State true or false.

i) Cube of any odd number is even. 

Ans: False. As multiplying an odd number three times will yield an odd number.

For example, the Cube of $5$ which is an odd number is \[25\], which is also an odd number.

ii) A perfect Cube does not end with two zeroes. 

Ans: True. Perfect Cube will always terminate with multiple of $3$ numbers of zeroes.

For example, the Cube of \[100\] is $1000000$ and there are $6$ zeros at the end of it.

iii) If square of a number ends with \[5\], then its Cube ends with \[25\]. 

Ans: False, it is not always certain that if the square of a number ends with \[5\], then its Cube will end with \[25\].

For examples, square of $55$ ends with 5, $3025$ but its Cube,$166375$ does not end with \[25\].

iv) There is no perfect Cube which ends with \[8\]. 

Ans: False, all the numbers having \[2\] at its unit digit place will have \[8\] in end as Cube.

v) The Cube of a two digit number may be a three digit number. 

Ans: False, as Cube of even smallest two digit number, $10$ is a four digit number,$1000$.

vi) The Cube of a two digit number may have seven or more digits. 

Ans: False, as Cube of even largest two digit number, $99$ is a six digit number, $970299$

vii) The Cube of a single digit number may be a single digit number. 

Ans: True, as a Cube of first two natural numbers, $1$ and $2$ are $1$ and $8$ respectively.

Overview of Deleted Syllabus for CBSE Class 8 Maths  Cubes and Cube Roots  

Class 8 Maths Chapter 6: Exercises Breakdown

Conclusion

Class 8 Maths Cube and Cube Roots  provides a solid foundation in understanding the properties of Cubes and Cube Roots . This chapter is important as it prepares students for more advanced mathematical concepts. Key points to focus on NCERT Class 8 Maths Chapter 6 include identifying perfect Cubes, calculating Cube Roots  using prime factorization, and recognizing patterns in Cubes. Vedantu's class 8 maths chapter 6 pdf NCERT Solutions offer clear and simple explanations, helping students solve exercises confidently. In previous years 3 questions from this chapter have been asked.

Other Study Material for CBSE Class 8 Maths Chapter 6

Chapter-Specific NCERT Solutions for Class 8  Maths

Given below are the chapter-wise NCERT Solutions for Class 8  Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.