# Class 8 RD Sharma Solutions – Chapter 4 Cubes and Cube Roots – Exercise 4.1 | Set 1

**Question 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)7Cube of 7 is

7 = 7× 7 × 7 = 343

(ii) 12Cube of 12 is

12 = 12× 12× 12 = 1728

(iii) 16Cube of 16 is

16 = 16× 16× 16 = 4096

(iv) 21Cube of 21 is

21 = 21 × 21 × 21 = 9261

(v) 40Cube of 40 is

40 = 40× 40× 40 = 64000

(vi) 55Cube of 55 is

55 = 55× 55× 55 = 166375

(vii) 100Cube of 100 is

100 = 100× 100× 100 = 1000000

(viii) 302Cube of 302 is

302 = 302× 302× 302 = 27543608

(ix) 301Cube of 301 is

301 = 301× 301× 301 = 27270901

**Question 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:**

Finding the Cube of natural numbers up to 10

1

^{3}= 1 × 1 × 1 = 12

^{3}= 2 × 2 × 2 = 83

^{3}= 3 × 3 × 3 = 274

^{3}= 4 × 4 × 4 = 645

^{3}= 5 × 5 × 5 = 1256

^{3}= 6 × 6 × 6 = 2167

^{3}= 7 × 7 × 7 = 3438

^{3}= 8 × 8 × 8 = 5129

^{3}= 9 × 9 × 9 = 72910

^{3}= 10 × 10 × 10 = 1000Hence, we conclude

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

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

**Question 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.

^{3}+ 2

^{3}+ 3

^{3}+…+ 9

^{3}by the above pattern.

**Solution:**

From the 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

**Question 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:**

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

So, let’s see how to find the cube of 3, 6, 9, 12 and 15

3

^{3}= 3 × 3 × 3 = 276

^{3}= 6 × 6 × 6 = 2169

^{3}= 9 × 9 × 9 = 72912

^{3}= 12 × 12 × 12 = 172815

^{3}= 15 × 15 × 15 = 3375As we can see that all the cubes are divisible by 27

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

**Question 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 **form** i.e. when divided by 3 it leaves the remainder 1’**

**Solution:**

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

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

4

^{3}= 4 × 4 × 4 = 647

^{3}= 7 × 7 × 7 = 34310

^{3}= 10 × 10 × 10 = 100013

^{3}= 13 × 13 × 13 = 219716

^{3}= 16 × 16 × 16 = 4096When all the above cubes are divided by ‘3’ leaves the remainder of 1.

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

**Question 6. Write the cubes 5 natural numbers of **form** 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**:

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

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

5

^{3}= 5 × 5 × 5 = 1258

^{3}= 8 × 8 × 8 = 51211

^{3}= 11 × 11 × 11 = 133114

^{3}= 14 × 14 × 14 = 274417

^{3}= 17 × 17 × 17 = 4913When all the above cubes are divided by ‘3’ leaves the remainder of 2

Hence, 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.

**Question 7. Write the cubes of 5 natural numbers of which are multiples of 7 and **verify** the following:**

**“The cube of a multiple of 7 is a multiple of 7**^{3}.

^{3}.

**Solution:**

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

So now, let us find the cube of 7, 14, 21, 28 and 35

7

^{3}= 7 × 7 × 7 = 34314

^{3}= 14 × 14 × 14 = 274421

^{3}= 21× 21× 21 = 926128

^{3}= 28 × 28 × 28 = 2195235

^{3}= 35 × 35 × 35 = 42875We can see that all the above cubes are multiples of 7

^{3}(343) as well.Hence, the statement“The cube of a multiple of 7 is a multiple of 7

^{3}is true.

**Question 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) 64Finding 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) 216Finding the factors of 216

216 = 2 × 2 × 2 × 3 × 3 × 3 = 2

^{3}× 3^{3}= 6^{3}Hence, it’s a perfect cube.

(iii) 243Finding the factors of 243

243 = 3 × 3 × 3 × 3 × 3 = 3

^{5}= 3^{3}× 3^{2}Hence, it’s not a perfect cube.

(iv)1000Finding the factors of 1000

1000 = 2 × 2 × 2 × 5 × 5 × 5 = 2

^{3}× 5^{3}= 10^{3}Hence, it’s a perfect cube.

(v)1728Finding the factors 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) 3087Finding the factors of 3087

3087 = 3 × 3 × 7 × 7 × 7 = 3

^{2}× 7^{3}Hence, it’s not a perfect cube.

(vii)4608Finding the factors of 4608

4608 = 2 × 2 × 3 × 113

Hence, it’s not a perfect cube.

(viii)106480Finding the factors of 106480

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

Hence, it’s not a perfect cube.

(ix) 166375Finding the factors of 166375

166375= 5 × 5 × 5 × 11 × 11 × 11 = 5

^{3}× 11^{3}= 55^{3}Hence, it’s a perfect cube.

(x)456533Finding the factors of 456533

456533= 11 × 11 × 11 × 7 × 7 × 7 = 11

^{3}× 7^{3}= 77^{3}Hence, it’s a perfect cube.

**Question 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.

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

(iv)4096 = 2^{12}= (2^{6})^{2}= 64^{2}As 64 is an even number. Its not a cube of odd natural number.

(v)32768 = 2^{15}= (2^{5})^{3}= 32^{3}As 32 is an even number. It’s not a cube of odd natural number.

(vi)6859 = 19 × 19 × 19 = 19^{3}It’s a cube of odd natural number.

**Question 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)675Finding the factors of 675.

675 = 3 × 3 × 3 × 5 × 5

= 3

^{3}× 5^{2}Hence, we need to multiply the product by 5.

(ii)1323Finding the factors of 1323

1323 = 3 × 3 × 3 × 7 × 7

= 3

^{3}× 7^{2}Hence, we need to multiply the product by 7 to make it a perfect cube.

(iii)2560Finding the factors of 2560

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

= 2

^{3}× 2^{3}× 2^{3}× 5Hence, we need to multiply the product by 5 × 5 = 25 to make it a perfect cube.

(iv)7803Finding the factors of 7803

7803 = 3 × 3 × 3 × 17 × 17

= 3

^{3}× 17^{2}Hence, we need to multiply the product by 17 to make it a perfect cube.

(v) 107811First find the factors of 107811

107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11

= 3

^{3}× 3 × 11^{3}Hence, we need to multiply the product by 3 × 3 = 9 to make it a perfect cube.

(vi)35721First find the factors of 35721

35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7

= 3

^{3}× 3^{3}× 7^{2}Hence, we need to multiply the product by 7 to make it a perfect cube.

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