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• RD Sharma Class 8 Solutions for Maths

# Class 8 RD Sharma Solutions – Chapter 4 Cubes and Cube Roots – Exercise 4.4 | Set 1

### (v) -32768

Solution:

(i) -125

The cube root of -125 will be ∛-125

= -∛125

= –∛5 × 5 × 5

= -5

Hence, the cube root of -125 is -5

(ii) -5832

The cube root of -5832 will be ∛-5832

= -∛2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

= –∛23 × 33 × 33 = -(2 × 3 × 3)

= -18

Hence, the cube root of -5832 is -18

(iii) -2744000

The cube root of -2744000 will be ∛-2744000

= -∛2 × 2 × 2 × 7 × 7 × 7 × 10 × 10 × 10

= -∛23 × 73 × 103 = -(2 × 7 × 10)

= -140

Hence, the cube root of -2744000 is -140

(iv) -753571

The cube root of –753571 will be ∛-753571

= -∛7 × 7 × 7 × 13 × 13 × 13

= -∛73 × 133

= -(7 × 13)

= -91

Hence, the cube root of –753571 is -91

(v) -32768

The cube root of -32768 will be ∛-32768

= -∛2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

= -∛23 × 23 × 23 × 23 × 23

= -(2 × 2 × 2 × 2 × 2)

= -32

Hence, the cube root of –32768 is -32

### (iv) ∛(-125 × -1000) = ∛-125 × ∛-1000

Solution:

(i) ∛27 × ∛64 = ∛(27 × 64)

In order to verify the equation, we must prove LHS = RHS

So, LHS = ∛27 × ∛64

= ∛3 × 3 × 3 × ∛4 × 4 × 4

= ∛(3 × 3 × 3) × (4 × 4 × 4) = 3 × 4 = 12

And, RHS = ∛(27 × 64)

= ∛(3 × 3 × 3 × 4 × 4 × 4) = 3 × 4 = 12

So, LHS = RHS = 12

Hence proved

(ii) ∛(64 × 729) = ∛64 × ∛729

In order to verify the equation, we must prove LHS = RHS

So, LHS = ∛ (64 × 729)

= ∛(4 × 4 × 4 × 9 × 9 × 9) = 4 × 9 = 36

And, RHS = ∛64 × ∛729

= ∛4 × 4 × 4 × ∛9 × 9 × 9

= ∛(4 × 4 × 4) × (9 × 9 × 9) = 4 × 9 = 36

So, LHS = RHS = 36

Hence proved

(iii) ∛(-125 × 216) = ∛-125 × ∛216

In order to verify the equation, we must prove LHS = RHS

So, LHS = ∛ (-125 × 216)

= ∛-5 × -5 × -5 × ∛2 × 2 × 2 × 3 × 3 × 3

= ∛(-5 × -5 × -5) × (2 × 2 × 2 × 3 × 3 × 3)

= -5 × 2 × 3 = -30

And, RHS = ∛-125 × ∛216

= ∛((-5 × -5 × -5 × 2 × 2 × 2 × 3 × 3 × 3)

= -5 × 2 × 3 = -30

So, LHS = RHS = -30

Hence proved

(iv) ∛(-125×-1000) = ∛-125 × ∛-1000

In order to verify the equation, we must prove LHS = RHS

So, LHS = ∛ (-125 × -1000)

= ∛-5 × -5 × -5 × ∛-10 × -10 × -10

= ∛(-5 × -5 × -5) × (-10 × -10 × -10) = -5 × -10 = 50

And, RHS = ∛-125 × ∛-1000

= ∛((-5 × -5 × -5 × -10 × -10 × -10)

= -5 × -10 = 50

So, LHS = RHS = 50

Hence proved

### (iv) -729 × -15625

Solution:

(i) 8×125

We know that ∛ab = ∛a × ∛b

So, ∛(8 × 125) = ∛8 × ∛125

= ∛(2 × 2 × 2) × ∛(5 × 5 × 5)

= 2 × 5 = 10

Hence, the cube root of 8 × 125 = 10

(ii) -1728 × 216

We know that ∛ab = ∛a × ∛b

So, ∛(-1728×216) = -∛1728 × ∛216

= -∛(2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3) × ∛(2 × 2 × 2 × 3 × 3 × 3)

= -(2 × 2 × 3 × 2 × 3) = -72

Hence, the cube root of -1728 × 216 = -72

(iii) -27 × 2744

We know that ∛ab = ∛a × ∛b

So, ∛(-27×2744) = -∛27 × ∛2744

= -∛(3 × 3 × 3) × ∛(2 × 2 × 2 × 7 × 7 × 7)

= -(3 × 2 × 7) = -42

Hence, the cube root of -27×2744 = -42

(iv) -729 × -15625

We know that ∛ab = ∛a × ∛b

So, ∛(-729 × -15625) = -∛729 × -∛15625

= -∛(3 × 3 × 3 × 3 × 3 × 3) × -∛(5 × 5 × 5 × 5 × 5 × 5)

= (3 × 3 × 5 × 5) = 225

Hence, the cube root of -729 × -15625 = 225

### (iv) 125 ∛a6 – ∛125a6

Solution:

(i) ∛(43 × 63)

We know that ∛(a × b) = ∛a × ∛b

So, ∛(43 × 63) = ∛43 × ∛63

= 4 × 6 = 24

Hence, ∛(43 × 63) = 24

(ii) ∛(8 × 17 × 17 × 17)

We know that ∛(a × b) = ∛a × ∛b

So, ∛ (8 × 17 × 17 × 17) = ∛8 × ∛17 × 17 × 17

= 2 × 17 = 34

Hence, ∛(8 × 17 × 17 × 17) = 34

(iii) ∛(700 × 2 × 49 × 5)

∛(700 × 2 × 49 × 5)

= ∛ 2 × 2 × 5 × 5 × 7 × 2 × 7 × 7 × 5

= ∛ 23 × 73 × 53

= 2 × 7 × 5 = 70

Hence, ∛(700 × 2 × 49 × 5) = 70

(iv) 125 ∛a6 – ∛125a6

125 ∛a6 – ∛125a6 = 125∛(a2)3 – ∛53 × (a2)3

= 125a2 – 5a2

= 120a2

Hence, 125 ∛a6 – ∛125a6 = 120a2

### (v) -39304/-42875

Solution :

(i) -125/729

We can write cube root of -125/729 as,

∛-125/ ∛729

= -(∛5 × 5 × 5)/(∛9 × 9 × 9)

= -5/9

Hence, the cube root of -125/729 is -5/9

(ii) 10648/12167

We can write cube root of 10648/12167 as,

∛10648/ ∛12167

= (∛2 × 2 × 2 × 11 × 11 × 11)/(∛23 × 23 × 23)

= (2 × 11)/ 23 = 22/23

Hence, the cube root of 10648/12167 is 22/23

(iii) -19683/24389

We can write cube root of -19683/24389 as,

∛-19683/ ∛24389

= -(∛3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3)/(∛29 × 29 × 29)

= -(3 × 3 × 3)/ 23 = -27/29

Hence, the cube root of19683/24389 is -27/29

(iv) 686/-3456

We can write cube root of 686/-3456 as,

∛686/ ∛-3456

= -(∛2 × 7 × 7 × 7)/(∛2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3)

= -7/ (2 × 2 × 3) = -7/12

Hence, the cube root of 686/-3456 is -7/12

(v) -39304/-42875

We can write the cube root of -39304/-42875 as,

∛-39304/ ∛-42875

= (∛2 × 2 × 2 × 17 × 17 × 17)/(∛5 × 5 × 5 × 7 × 7 × 7)

= (2 × 17)/ (5 × 7) = 34/35

Hence, the cube root of -39304/-42875 is 34/35

### (iv) 1.331

Solution :

(i) 0.001728 can be written as 1728/1000000

So cube root of 0.001728 can be derived as,

∛0.001728 = ∛1728 / ∛1000000

= ∛(2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3) / ∛(10 × 10 × 10 × 10 × 10 × 10)

= (2 × 2 × 3) / (10 × 10) = 12/100

= 0.12

Hence, the cube root of 0.001728 is 0.12

(ii) 0.003375 can be written as 3375/1000000

So cube root of 0.003375 can be derived as,

∛0.003375 = ∛3375 / ∛1000000

= ∛(3 × 3 × 3 × 5 × 5 × 5) / ∛(10 × 10 × 10 × 10 × 10 × 10)

= (3 × 5) / (10 × 10) = 15/100

= 0.15

Hence, the cube root of 0.003375 is 0.15

(iii) 0.001 can be written as 1/1000

So cube root of 0.001 can be derived as,

∛0.001 = ∛1 / ∛1000

= ∛(1 × 1 × 1) / ∛(10 × 10 × 10)

= 1/10 = 0.1

Hence, the cube root of 0.001 is 0.1

(iv) 1.331 can be written as 1331/1000

So cube root of 1.331 can be derived as,

∛1.331 = ∛1331 / ∛1000

= ∛(11 × 11 × 11) / ∛(10 × 10 × 10)

= 11/10 = 1.1

Hence, the cube root of 1.331 is 1.1

### (v) ∛(0.1 × 0.1 × 0.1 × 13 × 13 × 13)

Solution:

(i) Simplifying ∛27 + ∛0.008 + ∛0.064 we get,

= ∛3 × 3 × 3 + ∛0.2 × 0.2 × 0.2 + ∛0.4 × 0.4 × 0.4

= 3 + 0.2 + 0.4 = 3.6

Hence, ∛27 + ∛0.008 + ∛0.064 = 3.6

(ii) Simplifying ∛1000 + ∛0.008 – ∛0.125 we get,

= ∛10 × 10 × 10 + ∛0.2 × 0.2 × 0.2 – ∛0.5 × 0.5 × 0.5

= 10 + 0.2 – 0.5 = 9.7

Hence, ∛1000 + ∛0.008 – ∛0.125 = 9.7

(iii) Simplifying ∛(729/216) × 6/9 we get,

= ∛(9 × 9 × 9)/(6 × 6 × 6) × 6/9

= 9/6 × 6/9 = 1

Hence, ∛(729/216) × 6/9 = 1

(iv) Simplifying ∛(0.027/0.008) ÷ (0.09/0.04) – 1 we get,

= ∛(0.3 × 0.3 × 0.3)/(0.2 × 0.2 × 0.2) ÷ (0.3 × 0.3)/(0.2 × 0.2) – 1

= 0.3/0.2 ÷ 0.3/0.2 – 1 = 1 – 1 = 0

Hence, ∛(0.027/0.008) ÷ (0.09/0.04) – 1 = 0

(v) Simplifying ∛(0.1 × 0.1 × 0.1 × 13 × 13 × 13) we get,

= ∛(0.1 × 0.1 × 0.1 × 13 × 13 × 13) = 0.1 × 13

= 1.3

Hence, ∛(0.1 × 0.1 × 0.1 × 13 × 13 × 13) = 1.3

### (ii) ∛(-512)/∛(343) = ∛(-512/343)

Solution:

(i) ∛(729)/ ∛ (1000) = ∛(729/1000)

LHS = ∛(729)/ ∛(1000)

= ∛(9 × 9 × 9)/ ∛(10 × 10 × 10) = 9/10

RHS = ∛(729/1000)

= ∛(9 × 9 × 9/10 × 10 × 10) = 9/10

Since, LHS = RHS = 9/10

Hence, proved

(ii) ∛(-512)/ ∛(343) = ∛(-512/343)

LHS = ∛(-512)/ ∛(343)

= -∛(8 × 8 × 8)/ ∛(11 × 11 × 11) = -8/11

RHS = ∛(-512/343)

= -∛(8 × 8 × 8/(11 × 11 × 11) = -8/11

Since, LHS = RHS = -8/11

Hence, proved

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