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

Last updated date: 13th Sep 2024
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## NCERT Solution for Maths Class 8 Cubes and Cube Roots Chapter 6 - FREE PDF Download

NCERT Solutions for Class 8 Maths Cube and Cube Roots is an important resource that introduces students to the concept of Cubes and their corresponding Cube Roots. Understanding these concepts is crucial as they form the basis for more advanced mathematical studies and solving Maths Class 8 Chapter 6 solutions. This chapter focuses on the properties and patterns of Cubes and how to find Cube Roots of numbers using prime factorisation.

Table of Content
1. NCERT Solution for Maths Class 8 Cubes and Cube Roots Chapter 6 - FREE PDF Download
2. Glance on Maths Chapter 6  Class 8 - Cubes and Cube Root
3. Access Exercise wise NCERT Solutions for Chapter 6 Maths Class 8
4. Exercises Under NCERT Solutions for Class 8 Maths Chapter 6 Cubes and Cube Roots
5. Access NCERT Solutions for Class 8 Maths Chapter 6  – Cubes and Cube Roots
5.1Exercise 6.1
5.2Exercise 6.2
6. Overview of Deleted Syllabus for CBSE Class 8 Maths  Cubes and Cube Roots
7. Class 8 Maths Chapter 6: Exercises Breakdown
8. Other Study Material for CBSE Class 8 Maths Chapter 6
9. Chapter-Specific NCERT Solutions for Class 8  Maths
FAQs

Vedantu's class 8 maths chapter 6 pdf solutions provide detailed explanations and step-by-step methods to solve the exercises. The solutions are prepared by subject experts and are well-organized. You can download the NCERT Solutions for Class 8 Maths Chapter 6 PDF from Vedantu for free.

## Glance on Maths Chapter 6  Class 8 - Cubes and Cube Root

• Chapter 6 of Class 8 Maths deals with Cubes(perfect Cubes or Cube numbers),Cubes and their prime factors, Cube Root, finding Cube Roots  through the Prime Factorization method.

• A Cube of a number 𝑛 is obtained when the number is multiplied by itself three times.

• A perfect Cube is a number that can be expressed as the cube of an integer. It means the result of an integer multiplied by itself twice more.

• Finding the Cube Roots  of a number using prime factorization involves breaking down the number into its prime factors and then grouping these factors to determine the Cube Root.

• Steps to Find Cube Roots  by Prime Factorization:

1. Prime Factorization: Start by finding the prime factors of the given number. This means expressing the number as a product of prime numbers.

2. Group the Factors:  you have the prime factors, group them in sets of three identical factors.

3. Extract the Cube Root: For each group of three identical prime factors, take one factor out of the group.Multiply these factors together to get the Cube Roots of the original number.

• This article contains chapter notes, important questions, exemplar solutions, exercises and video links for Chapter 6 - Cubes and Cube Root, which you can download as PDFs.

• There are two exercises (6 fully solved questions) in class 8 maths chapter 6 Cubes and Cube Root.

## Access Exercise wise NCERT Solutions for Chapter 6 Maths Class 8

 Current Syllabus Exercises of Class 8 Maths Chapter Cubes and Cube Roots NCERT Solutions of Class 8 Maths Cubes and Cube Roots  Exercise 6.1 NCERT Solutions of Class 8 Maths Cubes and Cube Roots  Exercise 6.2

## Exercises Under 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

 2 216 2 108 2 54 3 27 3 9 3 3 1

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

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

 2 128 2 64 2 32 2 16 2 8 2 4 2 2 1

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

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

 2 1000 2 500 2 250 5 125 5 25 5 5 1

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

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.

 2 100 2 50 5 25 5 5 1

$100=2\times 2\times 5\times 5$

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

 2 46656 2 23328 2 11664 2 5832 2 2916 2 1458 3 729 3 243 3 81 3 27 3 9 3 3 1

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

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

 Chapter Dropped Topics Cubes and Cube Roots 7.3.2 Cube Roots of a Cube number

## Class 8 Maths Chapter 6: Exercises Breakdown

 Exercise Number of Questions Exercise 6.1 4 Questions and Solutions Exercise 6.2 2 Questions and Solutions

## 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.

## 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.

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

1. What is a cube root?

Cube root is the value that, when multiplied by itself three times, gives us a required number. To denote this using an example, 9 X 9 X 9 gives us 729. In this example, 9 is the cube root of 729. Similarly, 729 is the cube of 9.

Cube roots are represented symbolically using the symbol ∛. So, for the example given above, it is represented as ∛729 = 9.

2. What is the difference between a cube and a cube root?

When a number is multiplied by itself three times, the product so obtained is called the cube of the number. Now, you can say that the given number is the cube root of the product so obtained. For example, 27 is the cube of 3, as 3 x 3 x 3 gives 27; and 3 is the cube root of 27.

3. Does a cube root have two solutions like a square root?

No, cube roots do not have two solutions. A square root has two solutions, one positive and the other negative, but cube roots have only one solution. For example, the solutions for the square root of 16 are +4 and -4, that is, if you multiply +4 x +4 you will get 16 and if you multiply -4 x -4, the product will be 16. But there is a unique solution for every cube root. For example, if you multiply 3 x 3 x 3, the product will be +27, but if you multiply -3 x -3 x -3, the product will not be +27, instead, the product will be -27. Hence, +3 is the cube root of +27 and -3 is the cube root of -27.

4. Does a cube root have a sign?

Yes, the cube root of any given number will have the sign of the number itself. The cube root of any number can be positive or negative or zero. To find the cube root of a given number, you have to identify the prime factors of the number that are multiplied thrice to get the number. And if the given number has a positive or negative sign, then the cube root will have the same sign as the number itself.

For example, if the cube root of -216 has to be calculated, then the prime factors of 216 have to be found first, (2 x 2 x 2 x 3 x 3 x 3). Now keeping the negative sign outside the parentheses, it becomes - (2 x 2 x 2 x 3 x 3 x 3). Here, the prime factors can be divided into two groups of three common numbers, that is,

- {(2 x 2 x 2) x (3 x 3 x 3)}. So, the cube root is - (2 x 3) = -6. If the given number was +216, then the cube root would have been +6.

5. What are the different methods of finding the cube root of a given number?

Chapter 7 "Cube and Cube Roots" teaches us two main methods of finding cube roots of a given number:

• Cube root through prime factorization method

• Cube root when you know a number is a cube number

These methods are explained in great detail in Vedantu's NCERT Solutions for Chapter 7 of Class 8 Maths. Here, you can find the two methods mentioned above explained by our expert teachers using simple examples.

6. What is a Cube of a number in class 8 maths Cube and Cube Roots ?

A Cube of a number is the result of multiplying the number by itself twice more. For example, if the number is n, then its Cube is n×n×n. This concept is fundamental in understanding cubic numbers and their properties. Cubes are used in various mathematical applications and geometric calculations.

7. What is a perfect Cube in class 8 maths Cube and Cube Roots ?

A perfect Cube is a number that can be expressed as the Cube of an integer. For instance, 27 is a perfect Cube because it can be written as 3×3×3. Recognizing perfect Cubes is important for solving problems related to volume and other cubic measurements. Perfect Cubes also help in simplifying complex algebraic expressions.

8. How do you find the Cube Roots of a number in class 8 maths chapter 6?

The Cube Root of a number is found by determining which number, when multiplied by itself three times, gives the original number. For example, the Cube Root of 27 is 3 because 3×3×3=27. This concept is essential for solving equations involving cubic terms and for various real-world applications like volume calculations.

9. What is the prime factorization method for finding Cube Roots  in ncert class 8 maths chapter 6 ?

Prime factorization involves breaking down the number into prime factors, grouping them in sets of three, and taking one factor from each group. For example, to find the Cube Root of 64, you break it down to 2×2×2×2×2×2, group them into sets of three (2×2×2) and take one factor from each group, resulting in 2 3 =8. This method simplifies finding Cube Roots for large numbers.

10. What types of questions are in class 8 maths chapter 6 pdf solutions?

Exercise 6.2 includes questions on calculating Cube Roots using prime factorization. Students are required to break down numbers into their prime factors, group them accordingly, and then find the Cube Root. These questions help reinforce the understanding of Cube Roots and prime factorization, providing practical skills in number manipulation and problem-solving.

11. What is covered in ncert class 8 maths chapter 6?

Exercise 6.1 focuses on finding the Cubes of given numbers and identifying perfect Cubes. Students learn to calculate the Cube of a number and recognize which numbers are perfect Cubes. This exercise builds a foundation for understanding cubic relationships and prepares students for more complex operations involving Cubes and Cube Roots.

12. How many questions from this chapter appeared in previous exams from ncert class 8 maths chapter 6 ?

Three questions from this chapter appear in exams, focusing on both theoretical understanding and practical application. These questions typically involve finding Cubes, identifying perfect Cubes, and calculating Cube Roots using different methods. This highlights the importance of mastering these concepts for academic success and practical problem-solving.