NCERT Solutions for Class 8 Maths Chapter 7 - Cubes And Cube Roots

EXERCISE NO- 7.1

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 NO- 7.2

1. Find the cube root of each of the following numbers by prime factorisation method. 

(a) 64

Ans: Prime factorisation of 512 is

$64=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}$

Therefore,

$ \sqrt[3]{64}=2\times 2  $  

$ =4  $

(b) 512

Ans: Prime factorisation of 512 is 

$  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 $  

(c) 10648

Ans: Prime factorisation of 10648 is

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

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

(d) 27000

Ans: Prime factorisation of 27000

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

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

(e) 15625

Ans: Prime factorisation of 15625

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

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

(f) 13824

Ans: Prime factorisation of 13824

$  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} $  

 $  \sqrt[3]{13824}=2\times 2\times 2\times 3=24 $  

(g) 110592

Ans: Prime factorisation of 110592

$  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 $  

(h) 46656

Ans: Prime factorisation of 46656

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

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

(i) 175616

Ans: Prime factorisation of 175616

$  175616=\underline{2\times 2\times 2}\times \underline{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 $  

(j) 91125

Ans: Prime factorisation of 91125 

$  91125=\underline{3\times 3\times 3}\times \underline{3\times 3\times 3}\times \underline{5\times 5\times 5} $  

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

2. State true or false. 

(a) Cube of any odd number is even.

Ans: False.

Reason: When we find out the cube of an odd number, we will find an odd number as the result because the unit place digit of an odd number is odd and we are multiplying three odd numbers. Therefore, the product will again be an odd number. For example, the cube of 7 (i.e., an odd number) is 343, which is again an odd number.

(b) A perfect cube does not end with two zeroes. 

Ans: True.

Reason: Perfect cube will end with a certain number of zeroes that are always a perfect multiple of 3. For example, the cube of 10 is 1000 and there are 3 zeroes at the end of it. The cube of 100 is 1000000 and there are 6 zeroes at the end of it. 

(c) If a square of a number ends with 5, then its cube ends with 25.

Ans: False. 

Reason: It is not always necessary that if the square of a number ends with 5, then its cube will end with 25. For example, the square of 25 is 625, and 625 has its unit digit as 5. The cube of 25 is 15625. However, the square of 35 is 1225 and also has its unit place digit as 5 but the cube of 35 is 42875 which does not end with 25.

(d) There is no perfect cube that ends with 8.

Ans: False.

Reason: There are many cubes that will end with 8. The cubes of all the numbers having their unit place digit as 2 will end with 8. The cube of 12 is 1728 and the cube of 22 is 10648. 

(e) The cube of a two-digit number may be a three-digit number.

Ans: False. 

Reason: The smallest two-digit natural number is 10, and the cube of 10 is 1000 which has 4 digits in it. 

(f) The cube of a two-digit number may have seven or more digits.

Ans: False. 

Reason: The largest two-digit natural number is 99, and the cube of 99 is 970299 which has 6 digits in it. Therefore, the cube of any two-digit number cannot have 7 or more digits in it.

(g) The cube of a single-digit number may be a single-digit number. 

Ans: True.

Reason: as the cube of 1 and 2 are 1 and 8 respectively. 

3. You are told that 1331 is a perfect cube. Can you guess without factorization what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768 

Ans: Making groups of three digits starting from the rightmost digit of the number we get 2 groups, 1 and 331. Considering 331, The digit at its unit place is 1. We know that if the digit 1 is at the end of a perfect cube number, then its cube root will have its unit place digit as 1 only. Therefore, the unit place digit of the required cube root can be taken as 1. Taking the other group i.e., 1, The cube of 1 exactly matches with the number of the second group. Therefore, the tens digit of our cube root will be taken as the unit place of the smaller number whose cube is near to the number of the second group i.e., 1 itself. 1 will be taken as the tens place of the cube root of 1331. Hence, $\sqrt[3]{1331}=11$

The cube root of 4913 has to be calculated. Making groups of three digits starting from the rightmost digit of 4913 we get the groups like 4 and 913. Considering group 913, The number 913 ends with 3. It is known that if the digit 3 is at the end of a perfect cube number, then its cube root will have its unit place digit as 7 only. Therefore, the unit place digit of the required cube root is taken as 7. Taking the other group i.e., 4, We know that, 13 = 1 and 23 = 8 Also, 1 < 4 < 8 Therefore, 1 will be taken at the tens place of the required cube root. Thus, $\sqrt[3]{4913}=17$

The cube root of 12167 has to be calculated. Making groups of three digits starting from the rightmost digit of the number 12167, the groups are 12 and 167. Considering the group 167, 167 ends with 7. We know that if the digit 7 is at the end of a perfect cube number, then its cube root will have its unit place digit as 3 only. Therefore, the unit place digit of the required cube root can be taken as 3. Taking the other group i.e., 12, We know that, 23 = 8 and 33 = 27 Also, 8 < 12 < 27. 2 is smaller between 2 and 3. Therefore, 2 will be taken at the tens place of the required cube root. Thus, $\sqrt[3]{12167}=23$

The cube root of 32768 has to be calculated. Making groups of three digits starting from the rightmost digit of the number 32768, we get 2 get 32 and 768. Considering group 768, 768 ends with 8. We know that if the digit 8 is at the end of a perfect cube number, then its cube root will have its unit place digit as 2 only. Therefore, the unit place digit of the required cube root will be taken as 2. Taking the other group i.e., 32, We know that, 33 = 27 and 43 = 64 Also, 27 < 32 < 64 and 3 is smaller between 3 and 4. Therefore, 3 will be taken at the tens place of the required cube root. Thus, $\sqrt[3]{32768}=32$.

A Brief Overview of Class 8 Maths Chapter 7

NCERT Grade 8 Mathematics Chapter 7, Cubes and Cube Roots, allows the students to play with numbers, their multiples, and handle complex numbers easily. Cubes and Cube Roots teaches the students how to cube a number and find the cube root of a number with the help of simple formulae and examples for in-depth understanding.

Introduction to Cube and Cube Roots, Cubes, Some interesting patterns, Smallest multiple that is a perfect cube, Cube Roots, Cube root through prime factorisation method and Cube root of a cube number are some of the main topics covered in this chapter. Interesting explanations of concepts, activities, exercises, and simple language make this chapter interesting and fun to learn.

This chapter deals with an important topic to build the students' foundation while dealing with numbers and complex numerical operations. NCERT Grade 8 Mathematics Chapter 7 - Cubes and Cube Roots help the students to learn and master multiplication and division on the advanced level. The summary and assessments help the students to deal with the chapter and solve the sums given in the chapter. From the alpha to omega of dealing with cubes and cube roots, this chapter helps the students understand what a cube is, how it is derived, and other fundamentals. New concepts, interesting problems, and solved and unsolved examples in simple language make this chapter interesting and fun to learn.

Get 100 percent accurate NCERT Solutions for Class 8 Maths Chapter 7 (Cubes and Cube Roots) solved by expert Maths teachers. We provide step-by-step solutions for questions in the Class 8 Maths textbook as per CBSE Board guidelines from the latest NCERT book for Class 8 Maths.

The Topics and Sub-Topics in Chapter 7 Cubes and Cube Roots 

• Ex 7.1 - Introduction

• Ex 7.2 - Cubes

• Ex 7.2.1 - Some interesting patterns

• Ex 7.2.2 - Smallest multiple that is a perfect cube

• Ex 7.3 - Cube Roots

• Ex 7.3.1 - Cube root through prime factorisation method

• Ex 7.3.2 - Cube root of a cube number

We Cover All The Exercises of Chapter 7 Cubes and Cube Roots Given Below:-

Every question with its accurate solution of the  following exercise is covered in NCERT Solutions for Class 8 Maths Chapter 7 pdf to help you to prepare the chapter efficiently and score more marks.

Other than the given exercises, you should also practise all the solved examples given in the book to clear your concepts on Cubes and Cube Roots. You can also download the free PDF of Chapter 7 Cubes and Cube Roots and take the printout to keep it handy for your exam preparation. You can also Download Maths NCERT Solutions for Class 8 All Exercises to help you to revise complete Syllabus and score more marks in your examinations. Download NCERT Solutions for Class 8 Science and make use of it in your preparation. You can also find CBSE NCERT Solutions on our website to know more about the various subjects. The solutions are up-to-date and are sure to help in your academic journey.

Key Features of NCERT Solutions for Class 8 Maths Chapter 7

Some of the prominent features of NCERT Solutions of Class 8 Maths Chapter 7 are listed below.

• The NCERT solutions of Class 8 Maths Chapter 7 are curated by our subject-matter experts who have years of experience in this field to deliver you complete and precise solutions for all the exercise questions. 

• The answers are explained in a very simple language and step-by-step manner, which will help students strengthen their fundamentals of cube and cube roots. 

• Practising these questions will help improve your marks and boost your problem-solving skills.

• The solutions are available in free PDF format to download, and you can access this free PDF anytime and from anywhere.
  

Important Questions for Practice

1. Is 392 a perfect cube? If not, find the smallest natural number by which 392 should be multiplied so that the product obtained is a perfect cube.

2. Find the smallest number by which 128 must be divided to obtain a perfect cube.

3. Find the cube root of 13824 using the prime factorisation method.

4. Find the cube root of the 17576 using the estimation method.

5. Find the cube of 4.5

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