NCERT Solutions Class 8 Maths Chapter 6 Cubes and Cube Roots

Chapter 6 of Class 8 Mathematics, "Cubes and Cube Roots," is an essential topic that helps students build a strong math foundation. This chapter covers the basics of cube numbers and cube roots, including their properties and some interesting patterns between them. Understanding cube roots is important because they are the basis for many advanced calculations and algebraic expressions studied in higher classes. 

The NCERT Solutions for Class 8 Maths Chapter 6 provides clear explanations and step-by-step solutions to textbook exercises, making it easier for students to understand and solve problems related to finding cubes of big numbers by expansion techniques and other patterns. This blog is designed to help students master this chapter, boost their analytical skills, and gain confidence in math as they prepare for exams and future studies.

1.0Download NCERT Solutions for Class 8 Maths Chapter 6: Free PDF

Downloading the NCERT Solutions of Class 8 Chapter 6 PDF will help you understand the curriculum's key concepts. Below is the link to download the NCERT Solutions for Class 8 Maths Chapter 6 PDF

2.0Class 8 Maths Chapter 6: Breakdown of Exercises

3.0NCERT Solutions Class 8 Maths Chapter 6 Subtopics

Before understanding the NCERT solutions, it is important to know the subtopics covered under the chapter cubes and cube roots. This section explains the different subtopics covered in Chapter 6, "Cubes and Cube Roots." In the NCERT solutions, you will learn about:

• Perfect cubes or cube numbers
• Patterns of cubes
• Cube root through prime factorization method
• Cube root of numbers
• Cube root of a cube number

4.0NCERT Questions with Solutions for Class 8 Maths Chapter 6 - Detailed Solutions

Exercise : 6.1

• Which of the following numbers are not perfect cube? (i) 216 (ii) 128 (iii) 1000 (iv) 100 (v) 46656 Sol. (i) Resolving 216 into prime factors, we find that

$216=\underline{2\times 2\times 2}\times \underline{3\times 3\times 3}$ Clearly, the prime factors of 216 can be grouped into triples of equal factors and no factor is left over. $\therefore 216$ is a perfect cube (ii) Resolving 128 into prime factors, we find that

$128=\underline{2\times 2\times 2\times 2\times 2\times 2\times 2}$ Now, if we try to group together triples of equal factors, we are left with a single factor, 2. $\therefore 128$ is not a perfect cube. (iii) Resolving 1000 into prime factors, we find that

$1000=\underline{2\times 2\times 2\times 5\times 5\times 5}$ Clearly, the prime factors of 1000 can be grouped into triples of equal factors and no factor is left over. $\therefore 1000$ is a perfect cube. (iv) Resolving 100 into prime factors, we find that

$100=2\times 2\times 5\times 5$. Now, if we try to group together triples of equal factors, we are left with $2\times 2\times 5\times 5$ $\therefore 100$ is not a perfect cube. (v) Resolving 46656 into prime factors, we find that

$46656=\underline{2\times 2\times 2\times 2\times 2\times 2\times \underline{3\times 3\times 3}\times }$ $\underline{3\times 3\times 3}$ 46656 can be grouped into triples of equal factors and no factor is left over. $\therefore 46656$ is a perfect cube.

• 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

Sol. (i) Writing 243 as a product of prime factors, we have

$243=\underline{3\times 3\times 3}\times 3\times 3$ Clearly, to make it a perfect cube, it must be multiplied by 3 . (ii) Writing 256 as a product of prime factors, we have

$256=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2\times 2\times 2}$ Clearly, to make it a perfect cube, it must be multiplied by 2 . (iii) Writing 72 as a product of prime factors, we have

$72=\underline{2\times 2\times 2}\times 3\times 3$ Clearly, to make it a perfect cube, it must be multiplied by 3 . (iv) Writing 675 as a product of prime factors, we have

$675=\underline{3\times 3\times 3}\times 5\times 5$ Clearly, to make it a perfect cube, it must be multiplied by 5 . (v) Writing 100 as a product of prime factors, we have

$100=2\times 2\times 5\times 5$ Clearly, to make it a perfect cube, it must be multiplied by $2\times 5=10$.

• 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 Sol. (i) Writing 81 as a product of prime factors, we have

$81=\underline{3\times 3\times 3\times 3}$

Clearly, to make it a perfect cube, it must be divided by 3 . (ii) Writing 128 as a product of prime factors, we have

$128=\underline{2\times 2\times 2\times 2\times 2\times 2\times 2}$ Clearly to make it a perfect cube, it must be divided by 2 . (iii) Writing 135 as a product of prime factors, we have

$135=\underline{3\times 3\times 3}\times 5$ Clearly, to make it a perfect cube, it must be divided by 5 . (iv) Writing 192 as a product of prime factors, we have

$192=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times 3$ Clearly, to make it a perfect cube, it must be divided by 3 . (v) Writing 704 as a product of prime factors, we have

$704=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times 11$ Clearly, to make it a perfect cube, it must be divided by 11 .

• Parikshit makes a cuboid of plasticine of side $5\mathrm{cm},2\mathrm{cm},5\mathrm{cm}$. How many such cuboids will he need to form a cube? Sol. Volume of the cuboid $=5\times 2\times 5=2\times 5\times 5$ To make it a cube, we require $2\times 2\times 5$ Hence, volume of cube $=(2\times 5\times 5\times 2\times 2\times 5){\mathrm{cm}}^3$ or $(2\times 2\times 2\times 5\times 5\times 5){\mathrm{cm}}^3$. Number of cuboids $=\frac{\text{ Volume of cube }}{\text{ Volume of cuboid }}$ $=\frac{2\times 2\times 2\times 5\times 5\times 5}{2\times 5\times 5}=2\times 2\times 5$ $=20$ cuboids

Exercise: 6.2

• Find the cube root of each of the following numbers by prime factorisation method (i) 64 (ii) 512 (iii) 10648 (iv) 27000 (v) 15625 (vi) 13824 (vii) 110592 (viii) 46656 (ix) 175616 (x) 91125 Sol. (i) Resolving 64 into prime factors, we get

$64=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}$ $\therefore \sqrt[3]{64}=(2\times 2)=4$ (ii) Resolving 512 into prime factors, we get

$512=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times \underline{2\times 2\times 2}$ $\therefore \sqrt[3]{512}=(2\times 2\times 2)=8$ (iii) Resolving 10648 into prime factors, we get

$10648=\underline{2\times 2\times 2}\times \underline{11\times 11\times 11}$ $\therefore \sqrt[3]{10648}=(2\times 11)=22$ (iv) Resolving 27000 into prime factors, we get

$27000=\underline{2\times 2\times 2}\times \underline{3\times 3\times 3}\times \underline{5\times 5\times 5}$ $\therefore \sqrt[3]{27000}=(2\times 3\times 5)=30$ (v) Resolving 15625 into prime factors, we get

$15625=\underline{5\times 5\times 5}\times \underline{5\times 5\times 5}$ $\therefore \sqrt[3]{15625}=(5\times 5)=25$ (vi) Resolving 13824 into prime factors, we get

$13824=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times$ $\underline{3\times 3\times 3}$ $\therefore \sqrt[3]{13824}=(2\times 2\times 2\times 3)=24$ (vii) Resolving 110592 into prime factors, we get

$110592=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2\times \underline{2\times 2\times 2}}$ $\times \underline{2\times 2\times 2\times \underline{3\times 3\times 3}}$ $\therefore \sqrt[3]{110592}=(2\times 2\times 2\times 2\times 3)=48$ (viii) Resolving 46656 into prime factors, we get

$46656=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times \underline{3\times 3\times 3}\times$ $3\times 3\times 3$ $\therefore \sqrt[3]{46656}=(2\times 2\times 3\times 3)=36$ (ix) Resolving 175616 into prime factors, we get

$175616=\underline{2\times 2\times 2\times 2\times 2\times 2\times \underline{2\times 2\times 2}}$ $\times \underline{7\times 7\times 7}$ $\therefore \sqrt[3]{175616}=(2\times 2\times 2\times 7)=56$ (x) Resolving 91125 into prime factors, we get

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

• State true or false (i) Cube of any odd number is even. (ii) A perfect cube does not end with two zeros. (iii) If square of a number ends with 5 , then its cube ends with 25 . (iv) There is no perfect cube which ends with 8. (v) The cube of a two digit number may be a three digit number. (vi) The cube of two digit number may have seven or more digits. (vii) The cube of a single digit number may be a single digit number. Sol. (i) False, as $3^3=27$ is odd (ii) True, as $10^3=1000$ (iii) False, as $15^2=225$ and $15^3=3375$ (iv) False, as $8=2^3,1728=12^3$ etc. (v) False, as $10^3=1000$ (vi) False, as $99^3=970299$ (vii) True as $2^3=8$ and 8 is a single digit number.
• You are told that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root? Similarly, guess the cube roots of $4913,12167,32768$. Sol. (i) 1331 (1) Form group of three starting from the right most digit of 1331 . 1 331. In this case one group i.e., 331 has three digits whereas 1 has only one digit. (2) Take 331 . The digit 1 is at one's place. We take the one's place of the required cube root as 1 . (3) Take the other group, i.e. 1. Cube of 1 is 1 . So, take 1 as ten's place of the cube root of 1331 . Thus, $\sqrt[3]{1331}=11$ (ii) 4913 (1) Form group of three starting from the right most digit of 4913 . 4 913. In this case one group i.e., 913 has three digits whereas 4 has only one digit. (2) Take 913. The digit 3 is at one's place. We take the one's place of the required cube root as 7 . (3) Take the other group, i.e., 4. Cube of 1 is 1 and cube of 2 is 8.4 lies between $1\&8$. The smaller number among 1 and 2 is 1 . So, take 1 as ten's place of the cube root of 4913. Thus, $\sqrt[3]{4913}=17$ (iii) 12167 (1) Form group of three starting from the right most digit of 12167 . 12167 . In this case one group i.e., 167 has three digits whereas 12 has only two digits. (2) Take 167. The digit 7 is at its one's place. We take the one's place of the required cube root as 3 . (3) Take the other group, i.e., 12. Cube of 2 is 8 and cube of 3 is 27.12 lies between 8 & 27. The smaller number among 2 and 3 is 2 . So, take 2 as ten's place of the cube root of 12167 . Thus, $\sqrt[3]{12167}=23$ (iv) 32768 (1) Form group of three starting from the right most digit of 32768 . 32 768. In this case one group i.e., 768 has three digits whereas 32 has only two digits. (2) Take 768. The digit 8 is at its one's place. We take the one's place of the required cube root as 2 . (3) Take the other group, i.e., 32. Cube of 3 is 27 and cube of 4 is 64 . 32 lies between 27 & 64. The smaller number among 3 and 4 is 3 . So, take 3 as ten's place of the cube root of 32768 . Thus, $\sqrt[3]{32768}=32$