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NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 7 |

Chapter Name |
Cubes and Cube Roots |

Exercise |
Ex 7.1, Ex 7.2 |

Number of Questions Solved |
7 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots

### Chapter 7 Cubes and Cube Roots Exercise 7.1

**Question 1.**

Which of the following numbers are not perfect cubes ?

**(i)** 216

**(ii)** 128

**(iii)** 1000

**(iv)** 100

**(v)** 46656

**Solution:**

**(i)** **Resolving 216 into prime factors, we find that 2**

16 = 2x2x2 x 3x3x3

Clearly, the prime factors of 216 can be grouped into triples of equal factors and no factor is left over.

∴ 216 is a perfect cube.

**(ii) Resolving 128 into prime factors, we find that**

128 = 2x2x2 x 2x2x2 x 2

Now, if we try to group together triples of equal factors, we are left with a single factor, 2.

∴ 128 is not a perfect cube.

**(iii) Resolving 1000 into prime factors, we find that**

1000 = 10 x 10 x 10 = 2x5x2x5x2x5 = 2x2x2x 5x5x5

Clearly, the prime factors of 1000 can be grouped into triples of equal factors and no factor is left over.

∴ 1000 is a perfect cube.

**(iv) Resolving 100 into prime factors, we find that**

100 = 10×10=2x5x2x5=2x2x5x5

Clearly, the prime factors of 100 cannot be grouped into triples of equal

factors.

∴ 100 is not a perfect cube.

**(v) Resolving 46656 into prime factors, we find that**

46656 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x3 x 3x3x3

Clearly, the prime factors of 46656 can be grouped into prime factors and no factor is left over.

∴ 46656 is a perfect cube.

**Question 2.**

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

**Solution:**

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

243 = 3 x 3 x 3 x 3 x 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 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 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 = 2x2x2 x 3 x 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 = 3 x 3 x 3 x 5 x 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 x 2 x 5 x 5

Clearly, to make it a perfect cube, it must be multiplied 2 x 5, i.e., 10.

**Question 3.**

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

**(i)** 81

**(ii)** 128

**(iii)** 135

**(iv)** 192

**(v)** 704

**Solution:**

**(i)** Writing 81 as a product of prime factors, we have

81 = 3 x 3 x 3 x 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 = 2 x 2 x 2 x 2 x 2 x 2 x 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 = 3 x 3 x 3 x 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 = 2 x 2 x 2 x 2 x 2 x 2 x 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 = 2 x 2 x 2 x 2 x 2 x 2 x 11

Clearly, to make it a perfect cube, it must be divided by 11.

**Question 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?

**Solution:**

Volume of the cuboid = 5 x 2 x 5 = 2 x 5 x 5

To make it a cube, we require 2 x 2 x 5, i.e., 20 such cuboids.

### Chapter 7 Cubes and Cube Roots Exercise 7.2

**Question 1.**

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

**Solution:**

**(i)** Resolving 64 into prime factors, we get

64 = 2 x 2 x 2 x 2 x 2 x 2

∴

**(ii)** Resolving 512 into prime factors, we get

512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

∴

**(iii)** Resolving 10648 into prime factors, we get

10648 = 2 x 2 x 2 x 11 x 11 x 11

∴

**(iv)** Resolving 27000 into prime factors, we get

27000 = 1000 x 27

= 10 x 10 x 10 x 3 x 3 x 3 = 2 x 5 x 2 x 5 x 2 x 5 x 3 x 3 x 3

= 2 x 2 x 2 x 3 x 3 x 3 x 5 x 5 x 5

∴

**(v)** Resolving 15625 into prime factors, we get

15625 = 5 x 5 x 5 x 5 x 5 x5

∴

**(vi) **Resolving 13824 into prime factors, we get

13824 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3

∴

**(vii)** Resolving 110592 in prime factors, we get

110592 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3

∴

**(viii)** Resolving 46656 in prime factors, we get

46656 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3

∴

**(ix)** Resolving 175616 in prime factors, we get

175616 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 7 x 7 x 7 = (2 x 2 x 2 x 7)

= 56

**(x)** Resolving 91125 in prime factors, we get

91125 = 3 x 3 x 3 x 3 x 3 x 3 x 5 x 5 x 5

∴

**Question 2.**

**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 a two digit number may have seven or more digits.

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

**Solution:**

**(i)** False

**(ii)** True

**(iii)** False, as 15^{2} = 225 and 15^{3} = 3375

**(iv)** False, as 8 = 2^{3}, 1728 =12^{3}, etc.

**(v)** False

**(vi)** False, as 10^{3} = 1000, 99^{3} = 970299

**(vii)** True, as 1^{3} = 1, 2^{3} =8.

**Question 3.**

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.

**Solution:**

**For 1331 :**

Units digit of the cube root of 1331 is 1 as unit’s digit of the cube root of numbers ending in 1 is 1 . After striking three digits from the right of 1331, we get the number 1. Since 1^{3} = 1, so the ten’s digit of the cube root of given number is 1.

∴

**For 4913 :**

Units digit of the cube root of 4913 is 7 as unit’s digit of cube root of numbers ending in 3 is 7. After striking three digits from the right of 4913, we get the number 4. As 1^{3} =1 and 2^{3} =8, so 1^{3} < 4 < 2^{3}. Therefore, the ten’s digit of cube root of 4913 is 1.

∴

**For 12167 :**

Unit’s digit of the cube root of 12167 is 3 as unit’s digit of cube root of numbers ending in 7 is 3. After striking three digits from the right of 12167, we get the number 12. As 2^{3} =8 and 3^{3} =27, so 2^{3} < 12 < 3^{3}. So, the ten’s digit of the cube root of 12167 is 2.

∴

**For 32768 :**

Unit’s digit of the cube root of 32768 is 2 as unit’s digit of cube root of numbers ending in 8 is 2. After striking three digits from the right of 32768, we get the number 32. As 3^{3} =27 and 4^{3} = 64, so 3^{3} < 32< 4^{3}. So, the ten’s digit of the cube root of 32768 is 3.

∴

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