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

Exercise 4.4 | Set 1 of Chapter 4 in RD Sharma's Class 8 Mathematics textbook focuses on advanced applications of cubes and cube roots. This exercise set challenges students to apply their understanding of these concepts to more complex problems, often involving real-world scenarios. Students will work with larger numbers, multi-step problems, and questions that require a deeper level of analysis and problem-solving skills. The problems in this set are designed to reinforce previously learned concepts while introducing new ways to think about and apply knowledge of cubes and cube roots.

Question 1. Find the cube roots of each of the following integers:

(i) -125 

(ii) -5832

(iii) -2744000 

(iv) -753571

(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

Question 2. Show that:

(i) ∛27 × ∛64 = ∛(27 × 64)

(ii) ∛(64 × 729) = ∛64 × ∛729

(iii) ∛(-125 × 216) = ∛-125 × ∛216

(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

Question 3. Find the cube root of each of the following numbers:

(i) 8 × 125  

(ii) -1728 × 216

(iii) -27 × 2744 

(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

Question 4. Evaluate:

(i) ∛ (4³ × 6³)

(ii) ∛ (8 × 17 × 17 × 17)

(iii) ∛ (700 × 2 × 49 × 5)

(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

Question 5. Find the cube root of each of the following rational numbers:

(i) -125/729

(ii) 10648/12167

(iii) -19683/24389

(iv) 686/-3456

(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 of -19683/24389 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

Question 6. Find the cube root of each of the following rational numbers:

(i) 0.001728

(ii) 0.003375

(iii) 0.001

(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

Question 7. Evaluate each of the following:

(i) ∛27 + ∛0.008 + ∛0.064

(ii) ∛1000 + ∛0.008 - ∛0.125

(iii) ∛(729/216) × 6/9

(iv) ∛(0.027/0.008) ÷ √(0.09/0.04) - 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

Question 8. Show that:

(i) ∛(729)/∛(1000) = ∛(729/1000)

(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

Summary

Exercise 4.4 | Set 1 provides a comprehensive set of problems that challenge students to apply their knowledge of cubes and cube roots in various contexts. The questions range from straightforward calculations of cube roots to complex word problems involving real-world applications. Students practice working with large numbers, fractional and decimal cube roots, and algebraic expressions involving cubes. The problems also introduce concepts related to volume, surface area, and dimensional changes in cubical objects. This exercise set helps students develop critical thinking skills, mathematical reasoning, and the ability to translate real-world scenarios into mathematical equations involving cubes and cube roots.