# Question 1

**Find the cubes of the following numbers: (i) 7, (ii) 12, (iii) 21, (iv) 100, (v) 302**

**Solution:**

(i) (7)

^{3}= 7Ã—7Ã—7 = 343

(ii) (12)

^{3}= 12 Ã— 12 Ã— 12 = 1728

(iii) (21)

^{3}= 21 Ã— 21 Ã— 21 = 9621

(iv) (100)

^{3}= 100 Ã— 100 Ã— 100 = 1000000

(v) (302)^{3} = 302 Ã— 302 Ã— 302 = 27543608

# Question 2

**Write cubes of all natural numbers between 1 and 20 and verify theÂ following statements:**

**Â Â Â Â Â Â (a) Cubes of all odd natural numbers are odd.**

Â Â Â Â Â Â (b) Cubes of all even natural numbers are even.

Â Â Â Â Â Â (b) Cubes of all even natural numbers are even.

**Solution:**

(2)

^{3}= 8, (3)

^{3}= 27, (4)

^{3}= 64, (5)

^{3}= 125, (6)

^{3}= 216, ...Â (19)

^{3}=6859.

(a) Yes, cubes of all odd natural numbers are odd.Â

(b) Yes, cubes of all even natural numbers are even.

# Question 3

**Write cubes of 5 natural numbers which are multiples of 3 and verifyÂ theÂ following:**

**Â Â Â Â Â Â 'The cube of natural number, which is a multiple of 3 is a multiple of 27'.**

**Solution:**

(3)

^{3}= 3Ã— 3Ã— 3 =27

(6)

^{3}= 6 Ã— 6Ã— 6 =216

(9)

^{3}= 9 Ã— 9 Ã— 9 =729

(12)

^{3}= 12 Ã— 12 Ã— 12 =1728

(15)

^{3}= 15 Ã— 15 Ã— 15 =3375

Verification:

(3)

^{3}= 27 = 27 Ã— 1

(6)

^{3}= 216 = 27 Ã— 8

(9)

^{3}= 729 = 27 Ã— 27

(12)

^{3}= 1728 = 27 Ã— 64

(15)

^{3}= 3375 = 27 Ã— 125

.

^{.}. 'The cube of natural number, which is a multiple of 3 is a multiple of 27'.

**Solution:**

The 5 natural numbers which are of the form 3n + 1 (e.gÂ 4, 7, 10, â€¦) are as follows:

3 Ã—1 + 1 = 3 + 1 = 4

3 Ã—2 + 1 = 6 + 1 = 7

3 Ã—3 + 1 = 9 + 1 = 10

3 Ã—4 + 1 = 12 + 1 = 13

3 Ã—5 + 1 = 15 + 1 = 16Â

The cubes of 5 natural numbers which are of the form 3n + 1 (e.gÂ 4, 7, 10, â€¦) are as follows:

(4)

^{3}= 4 Ã— 4 Ã—4 = 64

(7)

^{3}= 7 Ã—7 Ã—7 = 343

(10)

^{3}= 10 Ã—10 Ã—10 = 1000

(13)

^{3}= 13 Ã—13 Ã—13 = 2197

(16)

^{3}= 16 Ã— 16 Ã— 16 = 4096Â

Verification:

64 = 3Â Ã— 21 + 1

343 = 3 Ã— 114 + 1

1000 = 3 Ã— 333 + 1

2197 = 3 Ã— 732 + 1

4096 = 3 Ã— 1365 + 1

.

^{.}.Â 'The cube of a natural number of the form 3n +1 is a natural number of the same form'.

# Question 5

**Write cubes of 5 natural numbers which are of the form 3n + 2Â (e.g. 5, 8, 11, â€¦)Â and verify the following: 'The cube of a natural number of the form 3n + 2Â is a natural number of theÂ same form'.**

**Solution:**

The 5 natural numbers which are of the form 3n + 2 (e.g 5, 8, 11 , â€¦) are as follows:

3 Ã—1 + 2 = 3 + 2 = 5

3 Ã—2 + 2 = 6 + 2 = 8

3 Ã—3 + 2 = 9 + 2 = 11

3 Ã— 4 + 2 = 12 + 2= 14

3 Ã—5 + 2 = 15 + 2 = 17Â

The cubes of 5 natural numbers which are of the form 3n + 2 (e.g 5, 8, 11, â€¦) are as follows:

(5)

^{3}= 5 Ã—5 Ã— 5 = 125

(8)

^{3}= 8 Ã— 8 Ã— 8 = 512

(11)

^{3}= 11 Ã— 11 Ã— 11 = 1331

(14)

^{3}= 14 Ã— 14 Ã— 14 = 2744

(17)

^{3}= 17 Ã— 17 Ã— 17 = 4913Â

Verification:

125Â = 3 Ã— 41 + 2

512Â = 3 Ã— 170 + 2

1331 = 3 Ã— 443 + 2

2744 = 3 Ã— 914 + 2

4913 = 3 Ã— 1637 + 2

.

^{.}. 'The cube of a natural number of the form 3n + 2 is a natural number of the same form'.

# Question 7

**What is the smallest number by which 392 must be multiplied so that theÂ product is a perfect cube ?**

**Solution:**

392Â = 2 Ã— 2 Ã— 2 Ã— 7 Ã— 7Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

7 occurs as a prime factor only twice.

Hence, 7 is the smallest number by which 392 must beÂ multiplied so that the product isÂ

a perfect cube.Â Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

# Question 8

**What is the smallest number by which 8640 must be divided so that theÂ quotientÂ is a perfect cube ?**

**Solution:**

8640 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 Ã— 5

5 occurs as a prime number only once.

Hence, 5 is the smallest number by which 8640 must beÂ

divided so that the quotient is a perfect cube.Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

# Question 10

**Find the cube root of:**

**Â Â Â Â Â Â Â Â (i) 343Â Â Â Â (ii) 1000Â Â Â (iii)2744Â Â Â (iv) 74088Â**

**Solution:**

(i) 343 = 7 Ã— 7 Ã— 7Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

.

^{.}. = 7

(ii) 1000 = 2 Ã— 2 Ã— 2 Ã— 5 Ã— 5 Ã— 5Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

.

^{.}. = 2Â Ã— 5 = 10

(iii) 2744 = 2 Ã— 2 Ã— 2 Ã— 7 Ã— 7 Ã— 7Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

.

^{.}. = 2Â Ã— 7 = 14

(iv) 74088 = 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 Ã— 7 Ã— 7 Ã— 7Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

.

^{.}. = 2Â Ã— 3Â Ã— 7 = 42

# Question 12

**Multiply 137592 by the smallest number so that the product is a perfectÂ cube.Â Also, find the cube root of the product.**

**Solution:**

137592 = 2Â´2Â´2Â´3Â´3Â´3Â´7Â´7Â´13

The number 7 and 13 should be multiplied once and twice

respectively so that the product is a perfect cube.

.

^{.}.Â The smallest number by which 137592 mustÂ

be multiplied = 7Â´13Â´13 = 1183Â

The required product = 137592 x 1183 =Â 2 x 2 x 2 x 3 x 3 x 3 x 7 x 7 x 13 x 7 x 13 x 13

= (2^{3} x 3^{3} x 7^{3} x 13^{3})

= (2 x 3 x 7 x 13)^{3}

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 546

# Question 13

**Divide the number 26244 by the smallest number so that the quotientÂ is aÂ perfect cube. Also, find the cube root of the quotient.Â**

**Solution:**

26244 = 2Ã—2Ã—3Ã—3Ã—3Ã—3Ã—3Ã—3Ã—3Ã—3Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

2x2x 3x 3 = 36 is the smallest number by which 26244 must be divided so that the quotient is a perfect cube.Â

Â Â Â Â Â Â Â

The required quotient is = 729

The cube root of the quotient =Â

**Solution:**

We know that, the volume of a cube = (side)

^{3}

The length of the side of a cube =

# Question 15

**Which of the following numbers are cubes of negative integers?**

**Â Â Â Â Â Â Â Â (a) -64Â Â Â Â Â (b) -2197Â Â Â Â Â Â (c) -1056Â Â Â Â Â Â Â (d) -3888**

**Solution:**

(a)Â Â Â 64 = 2x2x2x2x2x2Â

Â Â Â Â Â

.

^{.}.Â Â Â Â -64 is a cube of -4 a negative integer.

(b)Â Â 2197 = 13x13x13

Â Â Â Â

.

^{.}.Â Â -2197 is a cube of -13 a negative integer.Â

(c)Â Â 1056 = 2x2x2x2x2x3x11

In the above factorisation 2 x 3 x 3 x 11 x 11 remains after grouping in triplets. Therefore, 1056 is not a perfect cube.

Â Hence -1056 is not a cube of negative integer.

(d)Â Â 3888 = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3Â

In the above factorisation 2 x 3 x 3 remains after grouping in triplets. Therefore, 3888 is not a perfect cube.

Hence -3888 is not a cube of negative integer.

# Question 16

**Find the cube roots of:**

**Â Â Â Â Â Â Â Â (a) -125Â Â Â Â Â Â (b) -5832Â Â Â Â Â Â (c) -17576Â**

**Solution:**

(a)Â Â Â Â 125Â =Â 5x5x5

Â Â Â Â Â Â Â

(b)Â Â Â Â 5832 = 2x2x2x3x3x3x3x3x3

Â Â Â Â Â Â Â

(c)Â Â Â Â 17576Â =Â 2x2x2x13x13x13

Â Â Â Â Â Â Â

# Question 17

**Find the cube root of each of the following numbers:**

**Â Â Â Â Â Â Â Â 1. 8 Ã—64Â**

Â Â Â Â Â Â Â Â 2. (-216) Ã— 1728

Â Â Â Â Â Â Â Â 3. 27 Ã— (-2744)

Â Â Â Â Â Â Â Â 4. (-125)Ã—(-3375)

Â Â Â Â Â Â Â Â 5. -456533Â

Â Â Â Â Â Â Â Â 6. -5832000Â

Â Â Â Â Â Â Â Â 2. (-216) Ã— 1728

Â Â Â Â Â Â Â Â 3. 27 Ã— (-2744)

Â Â Â Â Â Â Â Â 4. (-125)Ã—(-3375)

Â Â Â Â Â Â Â Â 5. -456533Â

Â Â Â Â Â Â Â Â 6. -5832000Â

**Solution:**

(1)Â Â Â Â 8 x 64 = 2x2x2x2x2x2x2x2x2Â

Â Â Â Â Â

(2)Â Â Â 216Â x 1728 = 2x2x2x3x3x3x2x2x2x2x2x2x3x3x3

Â Â Â Â Â Â

(3)Â Â 27Â xÂ 2744 = 3x3x3x2x2x2x7x7x7

Â Â Â Â Â

(4)Â Â 125Â x 3375 = 5x5x5x3x3x3x5x5x5

Â Â Â Â Â

(5)Â Â Â 456533 = 7x7x7x11x11x11

Â Â Â Â Â Â

(6)Â Â Â 5832000Â =Â 5832Â x 1000 = 2x2x2x3x3x3x3x3x3x2x2x2x5x5x5

Â Â Â Â Â Â

# Question 18

**Find the cubes of the following by multiplication.**

**Â Â Â Â Â****Â (i)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â -4**

**Â Â Â Â Â Â****(ii)Â Â Â Â Â Â Â Â Â Â Â Â Â 23**

**Â Â Â Â Â Â****(iii)Â Â Â Â Â Â Â Â Â Â Â 3030**

**Â****Solution:**

(i)Â (-4)

^{3}=(4) x (4) x (4) = âˆ’64

(ii) (23)

^{3}=23 x 23 x 23 = 12167

(iii) (3030)

^{3}=3030 x 3030 x 3030 = 27818127000.

# Question 19

**Find the cube of the following rational numbers:**

**Â Â Â Â Â Â (i) 1.4**

**Solution:**

Â (i) (1.4)^{3} = 1.4 Ã— 1.4 Ã— 1.4 = 2.744.

# Question 20

**By what number would you multiply 231525 to make it a perfect cube?**

**Solution:**

The prime factorisation of 231525 is 5 Ã— 5 Ã— 3 Ã— 3 Ã— 3 Ã— 7 Ã— 7 Ã— 7.

The number that must be multiplied in order that the above product is a perfect cube is 5.

Therefore, Cube root of 231525 Ã— 5 is 5 Ã— 3 Ã— 7 = 105.