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**1. Find the cubes of the following numbers:(i) 7 (ii) 12**

**(iii) 16 (iv) 21**

**(v) 40 (vi) 55**

**(vii) 100 (viii) 302**

**(ix) 301**

**Solution:**

**(i)** 7

Cube of 7 is

7 = 7× 7 × 7 = 343

**(ii) **12

Cube of 12 is

12 = 12× 12× 12 = 1728

**(iii) **16

Cube of 16 is

16 = 16× 16× 16 = 4096

**(iv) **21

Cube of 21 is

21 = 21 × 21 × 21 = 9261

**(v) **40

Cube of 40 is

40 = 40× 40× 40 = 64000

**(vi)** 55

Cube of 55 is

55 = 55× 55× 55 = 166375

**(vii) **100

Cube of 100 is

100 = 100× 100× 100 = 1000000

**(viii) **302

Cube of 302 is

302 = 302× 302× 302 = 27543608

**(ix) **301

Cube of 301 is

301 = 301× 301× 301 = 27270901

**2.Write the cubes of all natural numbers between 1 and 10 and verify the following statements:(i) Cubes of all odd natural numbers are odd.(ii) Cubes of all even natural numbers are even.**

**Solutions:**

Firstly let us find the Cube of natural numbers up to 10

1^{3} = 1 × 1 × 1 = 1

2^{3} = 2 × 2 × 2 = 8

3^{3} = 3 × 3 × 3 = 27

4^{3} = 4 × 4 × 4 = 64

5^{3} = 5 × 5 × 5 = 125

6^{3} = 6 × 6 × 6 = 216

7^{3} = 7 × 7 × 7 = 343

8^{3} = 8 × 8 × 8 = 512

9^{3} = 9 × 9 × 9 = 729

10^{3} = 10 × 10 × 10 = 1000

∴ From the above results we can say that

(i) Cubes of all odd natural numbers are odd.

(ii) Cubes of all even natural numbers are even.

**3. Observe the following pattern:1 ^{3} = 1**

**1 ^{3} + 2^{3} = (1+2)^{2}**

**1 ^{3} + 2^{3} + 3^{3} = (1+2+3)^{2}**

Write the next three rows and calculate the value of 1^{3} + 2^{3} + 3^{3} +…+ 9^{3} by the above pattern.

**Solution:**

According to given pattern,

1^{3} + 2^{3} + 3^{3} +…+ 9^{3}

1^{3} + 2^{3} + 3^{3} +…+ n^{3} = (1+2+3+…+n)^{ 2}

So when n = 10

1^{3} + 2^{3} + 3^{3} +…+ 9^{3 }+ 10^{3} = (1+2+3+…+10)^{ 2}

= (55)^{2} = 55×55 = 3025

**4. Write the cubes of 5 natural numbers which are multiples of 3 and verify the followings:“The cube of a natural number which is a multiple of 3 is a multiple of 27’**

**Solution:**

We know that the first 5 natural numbers which are multiple of 3 are 3, 6, 9, 12 and 15

So now, let us find the cube of 3, 6, 9, 12 and 15

3^{3} = 3 × 3 × 3 = 27

6^{3} = 6 × 6 × 6 = 216

9^{3} = 9 × 9 × 9 = 729

12^{3} = 12 × 12 × 12 = 1728

15^{3} = 15 × 15 × 15 = 3375

We found that all the cubes are divisible by 27

∴ “The cube of a natural number which is a multiple of 3 is a multiple of 27’

**5.Write the cubes of 5 natural numbers which are of the form 3n + 1 (e.g. 4, 7, 10, …) and verify the following:“The cube of a natural number of the form 3n+1 is a natural number of the same from i.e. when divided by 3 it leaves the remainder 1’**

**Solution:**

We know that the first 5 natural numbers in the form of (3n + 1) are 4, 7, 10, 13 and 16

So now, let us find the cube of 4, 7, 10, 13 and 16

4^{3} = 4 × 4 × 4 = 64

7^{3} = 7 × 7 × 7 = 343

10^{3} = 10 × 10 × 10 = 1000

13^{3} = 13 × 13 × 13 = 2197

16^{3} = 16 × 16 × 16 = 4096

We found that all these cubes when divided by ‘3’ leaves remainder 1.

∴ the statement “The cube of a natural number of the form 3n+1 is a natural number of the same from i.e. when divided by 3 it leaves the remainder 1’ is true.

**6. Write the cubes 5 natural numbers of the from 3n+2(i.e.5,8,11….) and verify the following:“The cube of a natural number of the form 3n+2 is a natural number of the same form i.e. when it is dividend by 3 the remainder is 2’**

**Solution**:

We know that the first 5 natural numbers in the form (3n + 2) are 5, 8, 11, 14 and 17

So now, let us find the cubes of 5, 8, 11, 14 and 17

5^{3} = 5 × 5 × 5 = 125

8^{3} = 8 × 8 × 8 = 512

11^{3} = 11 × 11 × 11 = 1331

14^{3} = 14 × 14 × 14 = 2744

17^{3} = 17 × 17 × 17 = 4913

We found that all these cubes when divided by ‘3’ leaves remainder 2.

∴ the statement“The cube of a natural number of the form 3n+2 is a natural number of the same form i.e. when it is dividend by 3 the remainder is 2’ is true.

**7.Write the cubes of 5 natural numbers of which are multiples of 7 and verity the following:“The cube of a multiple of 7 is a multiple of 7 ^{3}.**

**Solution:**

The first 5 natural numbers which are multiple of 7 are 7, 14, 21, 28 and 35

So, the Cube of 7, 14, 21, 28 and 35

7^{3} = 7 × 7 × 7 = 343

14^{3} = 14 × 14 × 14 = 2744

21^{3} = 21× 21× 21 = 9261

28^{3} = 28 × 28 × 28 = 21952

35^{3} = 35 × 35 × 35 = 42875

We found that all these cubes are multiples of 7^{3}(343) as well.

∴The statement“The cube of a multiple of 7 is a multiple of 7^{3} is true.

**8. Which of the following are perfect cubes?(i) 64 (ii) 216(iii) 243 (iv) 1000(v) 1728 (vi) 3087(vii) 4608 (viii) 106480(ix) 166375 (x) 456533**

**Solution:**

**(i)** 64

First find the factors of 64

64 = 2 × 2 × 2 × 2 × 2 × 2 = 2^{6} = (2^{2})^{3} = 4^{3}

Hence, it’s a perfect cube.

**(ii)** 216

First find thefactors of 216

216 = 2 × 2 × 2 × 3 × 3 × 3 = 2^{3} × 3^{3} = 6^{3}

Hence, it’s a perfect cube.

**(iii)** 243

First find thefactors of 243

243 = 3 × 3 × 3 × 3 × 3 = 3^{5} = 3^{3} × 3^{2}

Hence, it’s not a perfect cube.

**(iv)** 1000

First find thefactors of 1000

1000 = 2 × 2 × 2 × 5 × 5 × 5 = 2^{3} × 5^{3} = 10^{3}

Hence, it’s a perfect cube.

**(v)** 1728

First find thefactors of 1728

1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = 2^{6} × 3^{3} = (4 × 3 )^{3} = 12^{3}

Hence, it’s a perfect cube.

**(vi)** 3087

First find thefactors of 3087

3087 = 3 × 3 × 7 × 7 × 7 = 3^{2} × 7^{3}

Hence, it’s not a perfect cube.

**(vii)** 4608

First find thefactors of 4608

4608 = 2 × 2 × 3 × 113

Hence, it’s not a perfect cube.

**(viii)** 106480

First find thefactors of 106480

106480 = 2 × 2 × 2 × 2 × 5 × 11 × 11 × 11

Hence, it’s not a perfect cube.

**(ix)** 166375

First find thefactors of 166375

166375= 5 × 5 × 5 × 11 × 11 × 11 = 5^{3} × 11^{3} = 55^{3}

Hence, it’s a perfect cube.

**(x)** 456533

First find thefactors of 456533

456533= 11 × 11 × 11 × 7 × 7 × 7 = 11^{3} × 7^{3} = 77^{3}

Hence, it’s a perfect cube.

**9. Which of the following are cubes of even natural numbers?216, 512, 729, 1000, 3375, 13824**

**Solution:**

**(i)** 216 = 2^{3} × 3^{3} = 6^{3}

It’s a cube of even natural number.

**(ii)** 512 = 2^{9} = (2^{3})^{3} = 8^{3}

It’s a cube of even natural number.

**(iii)** 729 = 3^{3} × 3^{3} = 9^{3}

It’s not a cube of even natural number.

**(iv)** 1000 = 10^{3}

It’s a cube of even natural number.

**(v)** 3375 = 3^{3} × 5^{3} = 15^{3}

It’s not a cube of even natural number.

**(vi)** 13824 = 2^{9 }× 3^{3} = (2^{3})^{3} × 3^{3} = 8^{3}×3^{3 }= 24^{3}

It’s a cube of even natural number.

**10. Which of the following are cubes of odd natural numbers?125, 343, 1728, 4096, 32768, 6859**

**Solution:**

**(i)** 125 = 5 × 5 × 5 × 5 = 5^{3}

It’s a cube of odd natural number.

**(ii)** 343 = 7 × 7 × 7 = 7^{3}

It’s a cube of odd natural number.

**(iii)** 1728 = 2^{6} × 3^{3} = 4^{3} × 3^{3} = 12^{3}

It’s not a cube of odd natural number. As 12 is even number.

**(iv)** 4096 = 2^{12} = (2^{6})^{2} = 64^{2}

Its not a cube of odd natural number. As 64 is an even number.

**(v)** 32768 = 2^{15} = (2^{5})^{3} = 32^{3}

It’s not a cube of odd natural number. As 32 is an even number.

**(vi)** 6859 = 19 × 19 × 19 = 19^{3}

It’s a cube of odd natural number.

**11. What is the smallest number by which the following numbers must be multiplied, so that the products are perfect cubes?(i) 675 (ii) 1323(iii) 2560 (iv) 7803(v) 107811 (vi) 35721**

**Solution:**

**(i)** 675

First find the factors of 675

675 = 3 × 3 × 3 × 5 × 5

= 3^{3} × 5^{2}

∴To make a perfect cube we need to multiply the product by 5.

**(ii)** 1323

First find the factors of 1323

1323 = 3 × 3 × 3 × 7 × 7

= 3^{3} × 7^{2}

∴To make a perfect cube we need to multiply the product by 7.

**(iii)** 2560

First find the factors of 2560

2560 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5

= 2^{3} × 2^{3} × 2^{3} × 5

∴To make a perfect cube we need to multiply the product by 5 × 5 = 25.

**(iv)** 7803

First find the factors of 7803

7803 = 3 × 3 × 3 × 17 × 17

= 3^{3} × 17^{2}

∴To make a perfect cube we need to multiply the product by 17.

**(v)** 107811

First find the factors of 107811

107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11

= 3^{3} × 3 × 11^{3}

∴To make a perfect cube we need to multiply the product by 3 × 3 = 9.

**(vi)** 35721

First find the factors of 35721

35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7

= 3^{3} × 3^{3} × 7^{2}

∴To make a perfect cube we need to multiply the product by 7.

**12. By which smallest number must the following numbers be divided so that the quotient is a perfect cube?(i) 675 (ii) 8640(iii) 1600 (iv) 8788(v) 7803 (vi) 107811(vii) 35721 (viii) 243000**

**Solution:**

**(i)** 675

First find the prime factors of 675

675 = 3 × 3 × 3 × 5 × 5

= 3^{3} × 5^{2}

Since 675 is not a perfect cube.

To make the quotient a perfect cube we divide it by 5^{2} = 25, which gives 27 as quotient where, 27 is a perfect cube.

∴ 25 is the required smallest number.

**(ii)** 8640

First find the prime factors of 8640

8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5

= 2^{3} × 2^{3} × 3^{3} × 5

Since 8640 is not a perfect cube.

To make the quotient a perfect cube we divide it by 5, which gives 1728 as quotient and we know that 1728 is a perfect cube.

∴5 is the required smallest number.

**(iii)** 1600

First find the prime factors of 1600

1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5

= 2^{3} × 2^{3} × 5^{2}

Since 1600 is not a perfect cube.

To make the quotient a perfect cube we divide it by 5^{2} = 25, which gives 64 as quotient and we know that 64 is a perfect cube

∴ 25 is the required smallest number.

**(iv)** 8788

First find the prime factors of 8788

8788 = 2 × 2 × 13 × 13 × 13

= 2^{2} × 13^{3}

Since 8788 is not a perfect cube.

To make the quotient a perfect cube we divide it by 4, which gives 2197 as quotient and we know that 2197 is a perfect cube

∴ 4 is the required smallest number.

**(v)** 7803

First find the prime factors of 7803

7803 = 3 × 3 × 3 × 17 × 17

= 3^{3} × 17^{2}

Since 7803 is not a perfect cube.

To make the quotient a perfect cube we divide it by 17^{2} = 289, which gives 27 as quotient and we know that 27 is a perfect cube

∴ 289 is the required smallest number.

**(vi)** 107811

First find the prime factors of 107811

107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11

= 3^{3} × 11^{3} × 3

Since 107811 is not a perfect cube.

To make the quotient a perfect cube we divide it by 3, which gives 35937 as quotient and we know that 35937 is a perfect cube.

∴ 3 is the required smallest number.

**(vii)** 35721

First find the prime factors of 35721

35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7

= 3^{3} × 3^{3} × 7^{2}

Since 35721 is not a perfect cube.

To make the quotient a perfect cube we divide it by 7^{2} = 49, which gives 729 as quotient and we know that 729 is a perfect cube

∴ 49 is the required smallest number.

**(viii)** 243000

First find the prime factors of 243000

243000 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5

= 2^{3} × 3^{3} × 5^{3} × 3^{2}

Since 243000 is not a perfect cube.

To make the quotient a perfect cube we divide it by 3^{2} = 9, which gives 27000 as quotient and we know that 27000 is a perfect cube

∴ 9 is the required smallest number.

**13. Prove that if a number is trebled then its cube is 27 time the cube of the given number.**

**Solution:**

Let us consider a number as a

So the cube of the assumed number is = a^{3}

Now, the number is trebled = 3 × a = 3a

So the cube of new number = (3a)^{ 3} = 27a^{3}

∴New cube is 27 times of the original cube.

Hence, proved.

**14. What happens to the cube of a number if the number is multiplied by(i) 3?(ii) 4?(iii) 5?**

**Solution:**

**(i)** 3?

Let us consider the number as a

So its cube will be = a^{3}

According to the question, the number is multiplied by 3

New number becomes = 3a

So the cube of new number will be = (3a)^{ 3} = 27a^{3}

Hence, number will become 27 times the cube of the number.

**(ii)** 4?

Let us consider the number as a

So its cube will be = a^{3}

According to the question, the number is multiplied by 4

New number becomes = 4a

So the cube of new number will be = (4a)^{ 3} = 64a^{3}

Hence, number will become 64 times the cube of the number.

**(iii)** 5?

Let us consider the number as a

So its cube will be = a^{3}

According to the question, the number is multiplied by 5

New number becomes = 5a

So the cube of new number will be = (5a)^{ 3} = 125a^{3}

Hence, number will become 125 times the cube of the number.

**15. Find the volume of a cube, one face of which has an area of 64m ^{2}.**

**Solution:**

We know that the given area of one face of cube = 64 m^{2}

Let the length of edge of cube be ‘a’ metre

a^{2} = 64

a = √ 64

= 8m

Now, volume of cube = a^{3}

a^{3} = 8^{3} = 8 × 8 × 8

= 512m^{3}

∴Volume of a cube is 512m^{3}

**16. Find the volume of a cube whose surface area is 384m ^{2}.**

**Solution:**

We know that the surface area of cube = 384 m^{2}

Let us consider the length of each edge of cube be ‘a’ meter

6a^{2} = 384

a^{2} = 384/6

= 64

a = √64

= 8m

Now, volume of cube = a^{3}

a^{3} = 8^{3} = 8 × 8 × 8

= 512m^{3}

∴ Volume of a cube is 512m^{3}

**17. Evaluate the following:(i) {(5 ^{2} + 12^{2})^{1/2}}^{3}(ii) {(6^{2} + 8^{2})^{1/2}}^{3}**

**Solution:**

(i) {(5^{2} + 12^{2})^{1/2}}^{3}

When simplified above equation we get,

{(25 + 144)^{1/2}}^{3}

{(169)^{1/2}}^{3}

{(13^{2})^{1/2}}^{3}

(13)^{3}

2197

(ii) {(6^{2} + 8^{2})^{1/2}}^{3}

When simplified above equation we get,

{(36 + 64)^{1/2}}^{3}

{(100)^{1/2}}^{3}

{(10^{2})^{1/2}}^{3}

(10)^{3}

1000

**18. Write the units digit of the cube of each of the following numbers:31, 109, 388, 4276, 5922, 77774, 44447, 125125125**

**Solution:**

**31**

To find unit digit of cube of a number we perform the cube of unit digit only.

Unit digit of 31 is 1

Cube of 1 = 1^{3} = 1

∴ Unit digit of cube of 31 is always 1

**109**

To find unit digit of cube of a number we perform the cube of unit digit only.

Unit digit of 109 is = 9

Cube of 9 = 9^{3} = 729

∴ Unit digit of cube of 109 is always 9

**388**

To find unit digit of cube of a number we perform the cube of unit digit only.

Unit digit of 388 is = 8

Cube of 8 = 8^{3} = 512

∴ Unit digit of cube of 388 is always 2

**4276**

To find unit digit of cube of a number we perform the cube of unit digit only.

Unit digit of 4276 is = 6

Cube of 6 = 6^{3} = 216

∴ Unit digit of cube of 4276 is always 6

**5922**

To find unit digit of cube of a number we perform the cube of unit digit only.

Unit digit of 5922 is = 2

Cube of 2 = 2^{3} = 8

∴ Unit digit of cube of 5922 is always 8

**77774**

To find unit digit of cube of a number we perform the cube of unit digit only.

Unit digit of 77774 is = 4

Cube of 4 = 4^{3} = 64

∴ Unit digit of cube of 77774 is always 4

**44447**

To find unit digit of cube of a number we perform the cube of unit digit only.

Unit digit of 44447 is = 7

Cube of 7 = 7^{3} = 343

∴ Unit digit of cube of 44447 is always 3

**125125125**

To find unit digit of cube of a number we perform the cube of unit digit only.

Unit digit of 125125125 is = 5

Cube of 5 = 5^{3} = 125

∴ Unit digit of cube of 125125125 is always 5

**19. Find the cubes of the following numbers by column method:(i) 35(ii) 56(iii) 72**

**Solution:**

**(i)** 35

We have, a = 3 and b = 5

Column I a^{3} | Column II 3×a^{2}×b | Column III 3×a×b^{2} | Column IV b^{3} |

3^{3} = 27 | 3×9×5 = 135 | 3×3×25 = 225 | 5^{3} = 125 |

+15 | +23 | +12 | 125 |

42 | 158 | 237 | |

42 | 8 | 7 | 5 |

∴ The cube of 35 is 42875

**(ii) **56

We have, a = 5 and b = 6

Column I a^{3} | Column II 3×a^{2}×b | Column III 3×a×b^{2} | Column IV b^{3} |

5^{3} = 125 | 3×25×6 = 450 | 3×5×36 = 540 | 6^{3} = 216 |

+50 | +56 | +21 | 126 |

175 | 506 | 561 | |

175 | 6 | 1 | 6 |

∴ The cube of 56 is 175616

**(iii) 72**

We have, a = 7 and b = 2

Column I a^{3} | Column II 3×a^{2}×b | Column III 3×a×b^{2} | Column IV b^{3} |

7^{3} = 343 | 3×49×2 = 294 | 3×7×4 = 84 | 2^{3} = 8 |

+30 | +8 | +0 | 8 |

373 | 302 | 84 | |

373 | 2 | 4 | 8 |

∴ The cube of 72 is 373248

**20. Which of the following numbers are not perfect cubes?(i) 64(ii) 216(iii) 243(iv) 1728**

**Solution:**

(i) 64

Firstly let us find the prime factors of 64

64 = 2 × 2 × 2 × 2 × 2 × 2

= 2^{3} × 2^{3}

= 4^{3}

Hence, it’s a perfect cube.

(ii) 216

Firstly let us find the prime factors of 216

216 = 2 × 2 × 2 × 3 × 3 × 3

= 2^{3} × 3^{3}

= 6^{3}

Hence, it’s a perfect cube.

(iii) 243

Firstly let us find the prime factors of 243

243 = 3 × 3 × 3 × 3 × 3

= 3^{3} × 3^{2}

Hence, it’s not a perfect cube.

(iv) 1728

Firstly let us find the prime factors of 1728

1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

= 2^{3} × 2^{3} × 3^{3}

= 12^{3}

Hence, it’s a perfect cube.

**21. For each of the non-perfect cubes in Q. No 20 find the smallest number by which it must be(a) Multiplied so that the product is a perfect cube.(b) Divided so that the quotient is a perfect cube.**

**Solution:**

Only non-perfect cube in previous question was = 243

**(a)** Multiplied so that the product is a perfect cube.

Firstly let us find the prime factors of 243

243 = 3 × 3 × 3 × 3 × 3 = 3^{3} × 3^{2}

Hence, to make it a perfect cube we should multiply it by 3.

**(b)** Divided so that the quotient is a perfect cube.

Firstly let us find the prime factors of 243

243 = 3 × 3 × 3 × 3 × 3 = 3^{3} × 3^{2}

Hence, to make it a perfect cube we have to divide it by 9.

**22. By taking three different, values of n verify the truth of the following statements:(i) If n is even, then n ^{3} is also even.(ii) If n is odd, then n^{3} is also odd.(ii) If n leaves remainder 1 when divided by 3, then n^{3} also leaves 1 as remainder when divided by 3.(iv) If a natural number n is of the form 3p+2 then n^{3} also a number of the same type.**

**Solution:**

**(i)** If n is even, then n^{3} is also even.

Let us consider three even natural numbers 2, 4, 6

So now, Cubes of 2, 4 and 6 are

2^{3} = 8

4^{3} = 64

6^{3} = 216

Hence, we can see that all cubes are even in nature.

Statement is verified.

**(ii)** If n is odd, then n^{3} is also odd.

Let us consider three odd natural numbers 3, 5, 7

So now, cubes of 3, 5 and 7 are

3^{3} = 27

5^{3} = 125

7^{3} = 343

Hence, we can see that all cubes are odd in nature.

Statement is verified.

**(iii)** If n leaves remainder 1 when divided by 3, then n^{3} also leaves 1 as remainder when divided by 3.

Let us consider three natural numbers of the form (3n+1) are 4, 7 and 10

So now, cube of 4, 7, 10 are

4^{3} = 64

7^{3} = 343

10^{3} = 1000

We can see that if we divide these numbers by 3, we get 1 as remainder in each case.

Hence, statement is verified.

**(iv)** If a natural number n is of the form 3p+2 then n^{3} also a number of the same type.

Let us consider three natural numbers of the form (3p+2) are 5, 8 and 11

So now, cube of 5, 8 and 10 are

5^{3} = 125

8^{3} = 512

11^{3} = 1331

Now, we try to write these cubes in form of (3p + 2)

125 = 3 × 41 + 2

512 = 3 × 170 + 2

1331 = 3 × 443 + 2

Hence, statement is verified.

**23. Write true (T) or false (F) for the following statements:(i) 392 is a perfect cube.(ii) 8640 is not a perfect cube.(iii) No cube can end with exactly two zeros.(iv) There is no perfect cube which ends in 4.(v) For an integer a, a**

^{3}is always greater than a

^{2}.(vi) If a and b are integers such that a

^{2}>b

^{2}, then a

^{3}>b

^{3}.(vii) If a divides b, then a

^{3}divides b

^{3}.(viii) If a

^{2}ends in 9, then a

^{3}ends in 7.(ix) If a

^{2}ends in an even number of zeros, then a

^{3}ends in 25.(x) If a

^{2}ends in an even number of zeros, then a

^{3}ends in an odd number of zeros.

**Solution:**

(i) 392 is a perfect cube.

Firstly let’s find the prime factors of 392 = 2 × 2 × 2 × 7 × 7 = 2^{3} × 7^{2}

Hence the statement is False.

(ii) 8640 is not a perfect cube.

Prime factors of 8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 = 2^{3} × 2^{3} × 3^{3} × 5

Hence the statement is True

(iii) No cube can end with exactly two zeros.

Statement is True.

Because a perfect cube always have zeros in multiple of 3.

(iv) There is no perfect cube which ends in 4.

We know 64 is a perfect cube = 4 × 4 × 4 and it ends with 4.

Hence the statement is False.

(v) For an integer a, a^{3} is always greater than a^{2}.

Statement is False.

Because in case of negative integers ,

(-2)^{2} = 4 and (-2)^{3} = -8

(vi) If a and b are integers such that a^{2}>b^{2}, then a^{3}>b^{3}.

Statement is False.

In case of negative integers,

(-5)^{2} > (-4)^{2} = 25 > 16

But, (-5)^{3} > (-4)^{3} = -125 > -64 is not true.

(vii) If a divides b, then a^{3} divides b^{3}.

Statement is True.

If a divides b

b/a = k, so b=ak

b^{3}/a^{3} = (ak)^{3}/a^{3} = a^{3}k^{3}/a^{3} = k^{3},

For each value of b and a its true.

(viii) If a^{2} ends in 9, then a^{3} ends in 7.

Statement is False.

Let a = 7

7^{2} = 49 and 7^{3} = 343

(ix) If a^{2} ends in an even number of zeros, then a^{3} ends in 25.

Statement is False.

Since, when a = 20

a^{2} = 20^{2} = 400 and a^{3} = 8000 (a^{3} doesn’t end with 25)

(x) If a^{2} ends in an even number of zeros, then a^{3} ends in an odd number of zeros.

Statement is False.

Since, when a = 100

a^{2} = 100^{2} = 10000 and a^{3} = 100^{3} = 1000000 (a^{3} doesn’t end with odd number of zeros)

**All Chapter RD Sharma Solutions For Class 8 Maths**

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