NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots have been prepared by our team of subject experts with an objective to help students in their finals. These books include the solution for all the questions given in the NCERT maths textbooks. Class 8 Maths Chapter 7 NCERT Solutions can be easily accessible by all the students either by downloading the PDF files or by studying online by visiting our website at BYJU’S. Those students who practicing questions from pdf, can gain more information about the chapter and can also have a quick review before their finals. The different types of NCERT solutions for class 8 maths teaches all the students about the various topics in the examination. Some of the different types of topics are Rational Numbers, Linear Equations in one variable and Practical Geometry.

### NCERT Solutions Class 8 Maths Chapter 7 Exercises

- NCERT Solutions Class 8 Maths Chapter 7 Cubes and Cube Roots Exercise 7.1
- NCERT Solutions Class 8 Maths Chapter 7 Cubes and Cube Roots Exercise 7.2

*Exercise 7.1 *

*Q1: *

*Mention the numbers that are not perfect cubes. *

*(A) 216*

*(B) 128*

*(C) 1000*

*(D) 100*

*(E) 46656*

**Solution:**

(A) 216

Prime factors of 216: 2x2x2x3x3x3

Here all the factors are in the groups of 3’s

Therefore, 216 is said to be a perfect cube number.

02 | 0216 |

02 | 0108 |

02 | 054 |

03 | 027 |

03 | 09 |

03 | 03 |

01 |

(B) 128

The prime factor of 128 = 2x2x2x2x2x2x2

Here one factor 2 does not appear in groups of 3

Hence, 128 is not a perfect cube.

02 | 0128 |

02 | 064 |

02 | 032 |

02 | 016 |

02 | 08 |

02 | 04 |

02 | 02 |

01 |

(C) 1000

The prime factors of 1000 = 2x2x2x 5x5x5

Here all the factors are in groups of 3

Hence, 1000 is said to be a perfect cube.

02 | 01000 |

02 | 0500 |

02 | 0250 |

05 | 0125 |

05 | 025 |

05 | 05 |

01 |

(D) 100

The prime factors of 100 is 2×2 x 5×5

Here all the factors do not appear in groups of 3.

Hence, 100 is not a perfect cube.

02 | 0100 |

02 | 050 |

05 | 025 |

05 | 05 |

01 |

(E) 46656

The prime factors of 46656 = 2x2x2x2x2x2x 3 x3x3x3x3x 3

Here all the factors are in groups of 3

Hence, 46656 is said to be a perfect cube.

02 | 046656 |

02 | 023328 |

02 | 011664 |

02 | 05832 |

02 | 02916 |

02 | 01458 |

03 | 0729 |

03 | 0243 |

03 | 081 |

03 | 027 |

03 | 09 |

03 | 03 |

01 |

*Q2 : *

*Find the smallest number when multiplied to obtain a perfect cube: *

* (A) 243*

*(B) 256*

*(C) 72 *

*(D) 675*

*(E) 100*

**Solution:**

(A) 243

The prime factors of 243 = 3x3x3x3x 3

Here 3 does not appear in groups of 3

Hence, For 243 to be a perfect cube it should be multiplied by 3.

03 | 0243 |

03 | 081 |

03 | 027 |

03 | 09 |

03 | 03 |

01 |

(B) 256

The prime factors of 256 is 2x2x2x2x2x 2 x2 x 2

Here one factor of 2 is required for it to make groups of 3.

Hence, for 256 to be a perfect cube it should be multiplied by 2.

02 | 0256 |

02 | 0128 |

02 | 064 |

02 | 032 |

02 | 016 |

02 | 08 |

02 | 04 |

02 | 02 |

01 |

(C) 72

The prime factors for 72 = 2 x2x2x 3x 3

Here the factor 3 does not appear in groups of 3

Hence, For 72 to be a perfect cube it should be multiplied by 3.

(D) 675

The prime factors for 675 = 3x3x3x 5×5

Here the factor 5 does not appear in groups of 3

Hence, for 675 to be a perfect cube it should be multiplied by 5.

03 | 0675 |

03 | 0225 |

03 | 075 |

05 | 025 |

05 | 05 |

01 |

(E) 100

The prime factors for 100 = 2x2x5x5

Here both the factors 2 and 5 are not in groups of 3

Hence, for 100 to be a perfect cube it should be multiplied by 2 and 5. ( i.e. 2 x 5 =10 )

02 | 0100 |

02 | 050 |

05 | 025 |

05 | 05 |

01 |

*Q3: *

*Find the smallest number by which when divided obtain a perfect cube. *

*(A) 81 *

*(B) 128*

*(C) 135*

*(D) 192 *

*(E) 704 *

**Solution:**

(A) 81

The prime factors for 81 = 3 x 3 x 3 x 3

Here, there is one factor of 3 which extra from the group of 3

Hence, for 81 to be a perfect cube it should be divided by 3.

03 | 081 |

03 | 027 |

03 | 09 |

03 | 03 |

01 |

(B) 128

The prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2

Here there is one factor of 2 which in not in the group of 3

Hence, for 128 to be a perfect cube then it should be divided by 2.

02 | 0128 |

02 | 064 |

02 | 032 |

02 | 016 |

02 | 08 |

02 | 04 |

02 | 02 |

01 |

(C) 135

The prime factors of 135 = 3 x 3 x 3 x 5

Here there is one factor of 5 which is not appearing with its group of 3.

Hence, for 135 to be a perfect cube it should be divided by 5.

03 | 0135 |

03 | 045 |

03 | 15 |

05 | 05 |

01 |

(D)192

The prime factors for 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3

Here there is one factor of 3 which does not appearing with its group of 3.

Hence for 192 to be a perfect cube then it should be divided by 3.

02 | 0192 |

02 | 096 |

02 | 048 |

02 | 024 |

02 | 012 |

02 | 06 |

03 | 03 |

01 |

(E) 704

The prime factor for 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11

Here there is one factor of 11 which is not appearing with its group of 3.

Hence for 704 to be a perfect cube it should be divided by 11.

02 | 0704 |

02 | 0352 |

02 | 0176 |

02 | 088 |

02 | 044 |

02 | 022 |

02 | 011 |

01 |

*Q4:*

*Reuben makes a cuboid of clay of sides 5 cm , 2 cm , 5 cm. If Reuben wants to form a cube how many such cuboids will be needed?*

**Solution:**

The numbers given: 5 x 2 x 5

Since the factors of 2 and 4 are both not in groups of 3.

Then, the number should be multiplied by 2 x 2 x 5 = 20 for it to be made a perfect cube.

Hence Reuben needs 20 cuboids.

*Exercise 7.2 *

*Q1 : *

*By the method of prime factorization find the cube root for the following. *

*(A) 64 *

*(B) 512 *

*(C) 10648*

*(D) 27000*

*(E) 15625 *

*(F) 13824 *

*(G) 110592 *

*(H) 46656*

*(I) 175616*

*(J) 91125*

**Solution:**

(A) 64

\(\sqrt[3]{64}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2}\\\sqrt[3]{64}=2\times 2\)

= 4

02 | 064 |

02 | 032 |

02 | 016 |

02 | 08 |

02 | 04 |

02 | 02 |

01 |

(B) 512

\(\sqrt[3]{512}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2}\)

= 2 x 2 x 2

= 8

02 | 0512 |

02 | 0256 |

02 | 0128 |

02 | 064 |

02 | 032 |

02 | 016 |

02 | 08 |

02 | 04 |

02 | 02 |

01 |

(C) 10648

= 2 x 11

=22

02 | 010648 |

02 | 05324 |

02 | 02662 |

011 | 01331 |

011 | 0121 |

011 | 011 |

01 |

(D) 27000

\(\sqrt[3]{27000}=\sqrt[3]{2\times 2\times 2\times 3\times 3\times 3\times 5\times 5\times 5}\)

=>2 x 3 x 5

=>30

02 | 027000 |

02 | 013500 |

02 | 06750 |

03 | 03375 |

03 | 01125 |

03 | 0375 |

05 | 0125 |

05 | 025 |

05 | 05 |

01 |

(E) 15625

\(\sqrt[3]{15625}=\sqrt[3]{5\times 5\times 5\times 5\times 5\times 5}\)

=> 5 x 5

=> 25

05 | 015625 |

05 | 03125 |

05 | 0625 |

05 | 0125 |

05 | 025 |

05 | 05 |

01 |

(F) 13824

\(\sqrt[3]{13824}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3}\)

=> 2 x 2 x 2 x 3

=> 24

02 | 13824 |

02 | 06912 |

02 | 03456 |

02 | 01728 |

02 | 0864 |

02 | 0432 |

02 | 0216 |

02 | 0108 |

02 | 054 |

03 | 27 |

03 | 09 |

03 | 03 |

01 |

(G) 110592

\(\sqrt[3]{110592}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2 \times3\times 3\times 3}\)

=> 2 x 2 x 2 x 2 x 3

=> 48

02 | 0110592 |

02 | 055296 |

02 | 027648 |

02 | 013824 |

02 | 06912 |

02 | 03456 |

02 | 01728 |

02 | 0864 |

02 | 0432 |

02 | 0216 |

02 | 0108 |

02 | 054 |

03 | 027 |

03 | 09 |

03 | 03 |

01 |

(H) 46656

\(\sqrt[3]{46656}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3\times 3\times 3\times 3}\)

=> 2 x 2 x 2 x 3 x 3 x 3

=> 36

(I) 175616

\(\sqrt[3]{175616}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 7\times 7\times 7}\)

=> 2 x 2 x 2 x 7

=> 56

02 | 0175616 |

02 | 087808 |

02 | 043904 |

02 | 021952 |

02 | 010976 |

02 | 05488 |

02 | 02744 |

02 | 01372 |

02 | 0686 |

07 | 0343 |

07 | 049 |

07 | 07 |

01 |

(J) 91125

\(\sqrt[3]{91125}=\sqrt[3]{3\times 3\times 3\times 3\times 3\times 3\times 5\times 5\times 5}\)

=> 3 x 3 x 5

=> 45

03 | 091125 |

03 | 010125 |

03 | 03375 |

03 | 01125 |

03 | 0375 |

05 | 0125 |

05 | 025 |

05 | 05 |

01 |

*Q2: *

*State whether the following is true of false:*

*(A) Any off number of a cube is even.*

*(B) When a number end with two zeros, it is never a perfect cube. *

*(C) If the square of a given number ends with 5 then its cube will end with 25. *

*(D) There is no number that ends with 8 which is a perfect cube.*

*(E) The cube of a given two digit number will always be a three digit number. *

*(F) The cube of a two digit number will have either seven or more digits.*

*(G) The cube of single digit number may also be a single digit number.*

**Solution:**

(A) The statement given is false.

Since, 1^{3 } = 1, 3^{3} = 27, 5^{3 }= 125, . . . . . . . . . . are all odd.

(B) The given statement is true.

Since, a perfect cube ends with three zeroes.

Eg. 10^{3} = 1000, 20^{3 }= 8000, 30^{3 } = 27,000 , . . . . . . . . . . . . so on.

(C) The given statement is false

Since, 5^{2} = 25, 5^{3} = 125 , 15^{2 }= 225, 15^{3 }= 3375 ( Did not end with 25)

(D) the given statement is false.

Since 12^{3} = 1728 [the number ends with 8]

22^{3} = 10648 [ the number ends with 8]

(E) The given statement is false

Since, 10^{3} = 1000 [Four digit number]

And 11^{3}= 1331 [four digit number]

(F) The statement is False.

Since 99^{3} = 970299 [Six digit number]

(G) the given statement is true

1^{3} = 1 [single digit]

2^{3 }= 8 [single digit]

*Q3 :*

*1331 is told to be a perfect cube. What are the factorization methods in which you can find its cube root? Similarly, find the cube roots for *

*(i)4913*

*(ii)12167*

*(iii)32768.*

**Solution:**

We know that 10^{3 }= 1000 and possible cute of 11^{3 } = 1331

Since, the cube of units digit is 1^{3 } = 1

Then, cube root of 1331 is 11

(i) 4913

We know that 7^{3} is 343

Next number that comes with 7 as the units place is 17^{3 } = 4913

Therefore the cube root of 4913 is 17

(ii) 12167

Since we know that 3^{3} = 27

Here in cube, the ones digit is 7

Now the next number with 3 In the ones digit is 13^{3} = 2197

And the next number with 3 in the ones digit is 23^{3 } = 12167

Hence the cube root of 12167 is 23

(iii) 32768

We know that 2^{3 }= 8

Here in the cube, the ones digit is 8

Now the next number with 2 in the ones digit is 12^{3 } = 1728

And the next number with 2 as the ones digit 22^{3 }= 10648

And the next number with 2 as the ones digit 32^{3} = 32768

Hence the cube root of 32768 is 32

Thus, these NCERT solutions are useful for training the students of class 8. These NCERT solutions for class 8 maths chapter 7 specifically deals with the different topics. Some of these different problems are perfect cubes, prime factorization etc. Thus, these are some of the various calculation metrics in chapter 8 . With the NCERT solutions for class 8 there are solutions for both maths and science. A student can download the PDFs below and go through the solutions for the problems. Thus these solutions are incredibly useful in providing the answers to the variety of questions asked. The solutions for math cubes and cube roots PDF can be downloaded easily and learnt whenever required. Thus, these are some of the Class 8 Maths Chapter 7 NCERT Solutions that are useful for learning these chapters.