# Rd Sharma Solutions Class 8 Chapter 4

## Rd Sharma Solutions Class 8 Chapter 4 Cubes and Cube Roots

Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of RD Sharma Class 8 Mathematics, Chapter 4, Cubes and Cube Roots. Here students can easily find Exercise wise solution for chapter 4, Cubes and Cube Roots. Students will find proper solutions for Exercise 4.1, 4.2, 4.3, 4.4 and 4.5 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily.

Cubes and Cube Roots Exercise 4.1 Solution :

Question no – (1)

Solution :

(i) Cube of 7

(7)³

7 = 7 × 7 × 7

= 343

Hence, the cube of 7 is 343.

(ii) Cube of 12

(12)³

= 12 = 12 × 12 × 12

= 1728

Thus, the cube of 12 is 1728.

(iii) Cube of 16

(16)³

= 16 = 16 × 16 × 16

= 4096
Thus, the cube of 16 is 4096

(iv) Cube of 21

(21)³

= 21 = 21 × 21 × 21

= 9261

Hence, the cube of 21 is 9261

(v) Cube of 40

(40)³

= 40 = 40 × 40 × 40

= 64000

∴ The cube of 40 is 64000

(vi) Cube of 55

(55)³

= 55 = 55 × 55 × 55

= 166375

Thus, the cube of 55 is 166375

(vii) Cube of 100

(100)³

= 100 = 100 × 100 × 100

= 1000000

Hence, the cube of 100 is 1000000

(viii) Cube of 302

(302)³

= 302 = 302 × 302 × 302

= 27543608

So, the cube of 302 is 27543608.

(ix) Cube of 301

(301)³

= 301 = 301 × 301 × 301

= 27270901

∴ The the cube of 301 is 27270901.

Question no – (2)

Solution :

1³ = 1 odd

2³ = 8 even

3³ = 27 odd

4³ = 64 even

5³ = 125 odd

6³ = 216 even

7³ = 343 odd

8³ = 512 even

9³ = 729 odd

10³ = 1000 even

∴ Both statements are verified.

Question no – (3)

Solution :

1³ + 2³ + 3³ + 4³ + 5³ + 6³ + 7³ + 8³ + 9³ + 10³

= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)²

= (55)²

= 3025

* 1³ + 2³ + 3³ + 4³

= (1 + 2 + 3 + 4)²

= (10)²

= 100

* 1³ + 2 ³ + 4³ + 4³ + 5³

= (1 + 2 + 3 + 4 + 5)

= (15)²

= 25

* 1³ + 2 ³ + 4³ + 4³ + 5³ + 6³

= (1 + 2 + 3 + 4 + 5 + 6)

= (21)²

= 441

Question no – (4)

Solution :

Cubes of 5 natural number –

3³ = 27

6³ = 216

9³ = 729

12³ = 1728

15³ = 3375

Those are multiple by ‘3’

Now the cubes as a multiple of ‘27’ are,

1 × 27 = 27

8 × 27 = 216

27 × 27 = 729

64 × 27 = 1728

125× 27 = 3375

Question no – (5)

Solution :

The natural number form (3n + 1),

(3 × 1 + 1) = 4

(3 × 2 + 1) = 7

(3 × 3 + 1) = 10

93 × 4 + 1) = 13

Those are divided by ‘3’ and leans reminded ‘1’

Question no – (6)

Solution :

(The ‘5’ natural number from (3n + 2),

(3 × 1 + 2) = 5

(3 × 2 + 2) = 8

(3 × 3 × 2) = 11

(3 × 4 + 20 = 14

(3 × 5 × 2) = 17

Those are divided by ‘3’ and leaves remainder ‘2’

Question no – (7)

Solution :

The ‘5’ natural number multiple by ‘7’,

14 = (14)³ = 2744

21 = (21)³ = 9261

28 = (28)³ = 21952

35 = (35)³ = 42875

42 = (42)³ 74088

Now, verify by 7³ = 343

2744/343 = 8

9261/343 = 27

21952/343 = 64

42875/343 = 125

74088/343 = 216

All are verify by 7³ = 343

Question no – (8)

Solution :

From the given numbers in the question the perfect cubes are,

(i) 64

= 2 × 2 × 2

= 2³

(ii) 216

= 6 × 6 × 6

= 6³

(iv) 1000

= 10 × 10 × 10

= 10³

(v) 1728

= 12 × 12 × 12

= 12³

(ix) 166375

= 55 × 55 × 55

= 55³

(x) 456533

= 77 × 77 × 77

= 77³

Question no – (9)

Solution :

Cube of even natural numbers are,

(i) 216 = 6³

(ii) 512 = 8³

(iii) 1000 = 10³

(iv) 13824 = 24³

Question no – (10)

Solution :

The cubes of odd natural number are,

125 = 5³

343 = 7³

6859 = 19³

Question no – (11)

Solution :

(i) 675

= (3 × 3 × 3) × (5 × 5)

5 should be multiplied.

(ii) 1323

= (3 × 3 × 3) × 7 × 7

7 should be multiplied.

(iii) 2560

= (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × 5

25 should be multiplied.

(iv) 7803

= (3 × 3 × 3) × 17 × 17

17 should be multiplied.

(v) 107811

= (3 × 3 × 3) × 3 × (11 × 11 × 11)

9 should be multiplied.

(vi) 35721

= (3 × 3 × 3) × (3 × 3 × 3) × 7 × 7

∴ 7 should be multiplied.

Question no – (12)

Solution :

(i) 675

= (3 × 3 × 3) × (5 × 5)

∴ 25 should be divided.

(ii) 8640

= (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3) × 5

∴ 5 should be divided.

(iii) 1600

= (2 × 2 × 2) × (2 × 2× 2) × (5 × 5)

∴ 25 should be divided.

(iv) 8788

= (2 × 2) × (13 × 13 × 13)

∴ 4 should be divided.

(v) 7803

= (3 × 3 × 3) × (17 × 17)

∴ 289 should be divided.

(vi) 107811

= (3 × 3 × 3) × 3 × (11 × 11 × 11)

∴ 3 should be divided.

(vii) 35721

= (3 × 3 × 3) × (3 × 3 × 3) × (7 × 7)

∴ 49 should be divided.

(viii) 243000

= (2 × 2 × 2) × (3 × 3 × 3) × (3 × 3) × (5 × 5 × 5)

9 should be divided.

Question no – (13)

Solution :

Let, the number is x,
∴ x = x³
When, number, = 3x
∴ 3x = (3x)³
= 27x³ …(Proved)

Question no – (14)

Solution :

(i) 3 = 27 times the cube of the number.

(ii) 4 = 64  times the cube of the number.

(iii) 5 = 125 times the cube of the number.

Question no – (15)

Solution :

Given, Area = 64 m³

one side of cube,

= (64)²

= 8 m

Volume,

= 8 × 8 × 8

= 512 m

Therefore, the volume of the cube will be 512 m.

Question no – (16)

Solution :

Given, Area = 384 m²

∴ One side of cube = (384) = 6a²

a² = 384/6 = 64

or, a = √64 = 8

∴ a = 8 cm

∴ Volume,

= a³ = 8³

= 512 cm³

Therefore, the volume of a cube will be 512 cm³.

Question no – (17)

Solution :

(i) [(5² + 12²)1/2}³

= {(25 + 144)1/2}³

= {√169}³

= (13)³

= 2197

(ii) {(6² + 8²)1/2}³

= {(36 + 64)1/2}³

= (√100)³

= (10)³

= 1000

Question no – (18)

Solution :

(i) 31 = 1 → 1

∴ The units digit cube of 31 is 1.

(ii) 109 = 9 → 9

∴ The units digit cube of 109 is 9.

(iii) 388 = 8 → 2

∴ The units digit cube of 388 is 2.

(iv) 833 = 3 → 7

∴ The units digit cube of 833 is 7.

(v) 4276 = 6 → 6

∴ The units digit cube of 4276 is 6.

(vi) 5922 = 2 → 8

∴ The units digit cube of 5922 is 8.

(vii) 77774 = 4 → 4

∴ The units digit cube of 77774 is 4.

(viii) 44447 = 4 → 3

∴ The units digit cube of 44447 is 3.

(ix) 125125125 = 5 → 5

∴ The units digit cube of 125125125 is 5.

Question no – (19)

Solution :

(i) 35

= a = 3,  b = 5

 a³ 3a²b 3a²b b³ 3³ 3 × 3² × 5 3 × 5² × 3 5³ 27 135 225 125 15 23 12 42 8 7 5

(35)³ = 42875

(ii) 56

= a = 5,  b = 6

 a³ 3a²b 3a²b b³ 5³ 3 × 5² × 6 3 × 6² × 5 6³ 125 450 540 216 50 56 21 175 506 561 6

∴  (56)² = 175616

(iii) 72

= a = 7,  b = 2

 a³ 3a²b 3a²b b³ 7³ 3 × 7² × 2 3 × 2² × 7 2³ 343 294 + 8 84 8 30 302 373 2 4 8

∴ (72)³ = 372348

Question no – (20)

Solution :

The number which is not perfect cube is 243

243 = (3 × 3 × 3) × 3

Therefore, 243 is not a perfect cube.

Question no – (21)

Solution :

(a) multiplied so that the product is a perfect cube.

Solution : 9 should be multiplied to make a perfect cube.

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

Solution : 3 should be multiplied to take a perfect cube.

Question no – (22)

Solution :

(i) If n is even, then n³ is also even.

When = n = 2 = 2³ = 8 also even.

(ii) if n is odd, then n³ is also odd.

When = n = 3 = 3³ = 27 also odd.

Question no – (23)

Solution :

(i) 392 is a perfect cube – False.

(ii) 8640 is not a perfect cube – True

(iii) No cube can end with exactly two zeros – True

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

(v) For an integer a, a³ is always greater than a² – False.

(vi) If a and b are integers such that a² > b², then a³ > b³ – False.

(vii) If a divides b, then a³ divides b³ – True.

(viii) If a² ends in 9, then a³ ends in 7 – False.

(ix) If a² ends in 5, then a³ ends in 25 – False.

(x) If a² ends in an number of zeros, then a³ ends in an odd number of zeros – False.

Cubes and Cube Roots Exercise 4.2 Solution :

Question no – (1)

Solution :

(i) Given number, (-11)

= (-11)³

= -11 × -11 × -11

= -1331

Thus, the cube of -11 is -1331.

(ii) (-12)

= (-12)³

= (-12 × -12 × -12)

= -1728

Thus, the cube of -12 is -1728.

(iii) (- 21)

= (-21)³

= (-21 × -21 × -21)

= -9261

Therefore, The cube of -21 is -9261

Question no – (2)

Solution :

(i) -64

= -4³

64 is a cube of negative integer -4.

(iii) -2197

= 13³

-2197 is a cube of negative integer 13³.

(iv) -2744

= 14³

-2744 is a cube of negative integer 14³.

(v) -42875

= 35³

-42875 is a cube of negative integer 35³.

Question no – (4)

Solution :

(i) 7/9

(7/9)³ = 7 × 7 × 7/9 × 9 × 9

= 343/729

So, the the cube of 7/9 is = 343/729

(ii) -8/11

= (-8/11)³

= 8 × 8 × 8/11 × 11 × 11

= -512/1331

The cube of -8/11 of -512/1331.

(iii) 12/7

(12/7)³

= 12 × 12 × 12/7 × 7 × 7

= 1728/343

Hence, the cube of 12/7 is 1728/343.

(iv) -13/8

(-13/8)³

= 13 × 13 × 13/8 × 8 × 8

= -2197/512

Therefore, the cube of -13/8 is -2197/512.

(v) 2 2/5 = (12/5)

(12/5)³

= 12 × 12 × 12/ 5 × 5 × 5

= 1728/125

Hence, the cube of 2 2/5 is 1728/125.

(vi) 3 1/4 = 13/4

(13/4)³

= 13×13×13/4×4×4

= 2197/64

Thus, the cube of 3 1/4 is 2197/64.

(vii) 0.3

= (3/10)³

= 3×3×3/10×10×10

= 27/1000

= 0.027

Thus, the cube of 0.3 is 0.027.

(viii) 1.5

∴ (15/10)³

= 15×15×15/10×10×10

= 3375/1000

= 3.375

Thus, the cube of 1.5 is 3.375.

(ix) 0.08

= (8/100)³

= 8 × 8 × 8/100 × 100 × 100

= 512/100000

= 0.000512

Thus, the cube of 0.08 is 0.000512.

(x) 2.1

= (21/10)³

= 21×21×21/10×10×10

= 9261/1000

= 9.261

Hence, the cube of 2.1 is 9.261.

Question no – (5)

Solution :

(i) 27/64

= 3 × 3 × 3/ 4 × 4 × 4

= 3/4

(ii) 125/128

= 5 × 5 × 5/2 × 2 × 2 × 2 × 2 × 2

= 125/64

(iii) 0.001331

= 1331/1000000

= 11 × 11 × 11/100 × 100 × 100

= 11/100

= 0.11

(iv) 0.04

= 4/100

= 2 × 2/10 × 10

= 2/10 0.2

(i) and (ii) are rational numbers.

Cubes and Cube Roots Exercise 4.3 Solution :

Question no – (1)

Solution :

(i) 64

= ∛64 = 4

The cube root of 64 is 4.

(ii) 512

= ∛512 = 8

The cube root of 512 is 8.

(iii) 1728

= ∛1728 = 12

The cube root of 1728 is 12.

Question no – (2)

Solution :

(i) 130

= 130 – 1 = 129

129 – 7 = 122

122 – 19 = 103

103 – 37 = 66

66 – 61 = 5

Next number to be subtracted is 91, which greater than 5.

It is a not perfect cube.

(ii) 345

= 345 – 244

344 – 7 = 337

337 – 19 = 318

318 – 37 = 281

281 – 61 = 220

220 – 91 = 129

129 – 127 = 2

Next number subtracted is 169 which is greater than ‘2’

It is not perfect cube

(iii) 792

= 792 – 1 = 791

791 – 7 = 784

784 – 19 = 765

765 – 37 = 728

728 – 61 = 667

667 – 91 = 576

576 – 127 = 449

449 – 169 = 280

280 – 217 = 63

Next number to be subtracted is 271. which in greater than 63

729 is not a perfect cube

(iv) 1331

= 1331 -1 = 1330

1331 – 7 = 1323

1331 – 19 = 1304

1331 – 37 = 1267

1331 – 61 = 1206

1331 – 91 = 1115

1331 – 127 = 988

1331 – 169 = 819

1331 – 217 = 602

1331 – 271 = 331

1331 – 331 = 0

It is a perfect cube.

Question no – (3)

Solution :

(i) 130

= 5 should be subtracted

130 – 5

= 125

= 5³

(ii) 345

= 2 should be subtracted

345 – 2

= 343

= 7³

(iii) 729 = 63

= 63 should be subtracted.

729 – 63

= 729

= 9³

Question no –  (4)

Solution :

(i) 343

= ∛343

= ∛7 × 7 × 7

= 7

Thus, the cube root of 343 is 7.

(ii) 2744

= ∛2744

= ∛14 ×14 × 14

= 14

Thus, the cube root of 2744 is 14.

(iii) 4913

= ∛4913

= ∛17 × 17 × 17

= 17

Hence, the cube root of 4913 is 17

(iv) 1728

= ∛1728

= ∛12 × 12 × 12

= 12

Thus, the cube root of 1728 is 12.

(v) 35937

= ∛35937

= ∛33 × 33 × 33

= 33

Thus, the cube root of 35937 is 33.

(vi) 17576

= ∛17576

= ∛26 × 26 × 26

= 26

Hence, the cube root of 17576 is 26.

(vii) 134217728

= ∛134217728

= ∛512 × 512 × 512

= 512

Therefore, the cube root of 134217728 is 512.

(viii) 48228544

= ∛48228544

= ∛364 × 364 × 364

= 364

The cube root of 48228544 is 364.

(ix) 74088000

= ∛74088000

= ∛420 × 420 × 420

= 420

Therefore, the cube root of 74088000 is 420

(x) 157464

= ∛157464

= ∛54 × 54 × 54

= 54

Hence, the cube root of 157464 is 54

(xi) 1157625

= ∛1157625

= ∛105 × 105 × 105

= 105

Thus, the cube root of 1157625 is 105

(xii) 33698267

= ∛33698267

= ∛323 × 323 × 323

= 323

The cube root of 33698267 is 323.

Question no – (5)

Solution :

Given number, 3600

= 3 × 3 × 2 × 2 × 2 × 2 × 5 × 5

a perfect cube has multiplied of 3.

3 × 2 × 2 × 5

= 60 to make a perfect cube.

3600 × 60

= 21600 in the required.

∛216000

= 3√216000

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

= 3 × 2 × 2 × 5

= 60

Therefore, the cube root of the product will be 60.

Question no – (6)

Solution :

Given number in the question, 210125

= 5 × 5 × 5 × 41 × 41

41 should be multiplied to make a perfect also

The required number,

= 210125 × 41

= 8615125

∛8615125

= ∛5 × 5 × 5 × 41 × 41 × 41

= 205

Question no – (7)

Solution :

Given number, 8192

= (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × 2

2 should be divided to make a perfect cube.

The required number,

= 8192/2

= 4096

∛4096

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

= 2 × 2 × 2 × 2

= 16

Question no – (8)

Solution :

Let, common factor = x

According to question,

x³ + (2x)³ + (3x)³ = 98784

or, x³ + x³ + 27x³ = 98784

or, 36 x³ = 98784

or, 98784/36 = 2272

or, x = ∛2272 = 14

There numbers are,

x = 14,

2x = 14 × 2 = 28

3x = 14 × 3 = 48

Question no – (9)

Solution :

As per the given question,

Vol of cube = 9261000 m³

One side of cube,

= 3√9261000

= 210 m

Thus, the side of the cube will be 210 m.

Cubes and Cube Roots Exercise 4.4 Solution :

Question no – (1)

Solution :

(i) -125

∛ -125 = -5

The cube roots of -125 is -5.

(ii) –5832

∛- 5832 = -18

The cube roots of –5832 is -18.

(iii) -2744000

∛- 274000 = -140

The cube roots of -2744000 is -140.

(iv) -753571

∛- 753571 = -91

The cube roots of -753571 is -91.

(v) -32768

∛ -32768 = -32

The cube roots of -32768 is -32.

Question no – (2)

Solution :

(i) ∛27 × ∛64 = ∛27 × 64

∴ L.H.S, ∛27 × ∛64

= 3 × 4 = 12

R.H.S, ∛27 × 64

= ∛1728

= 12

L.H.S = R.H.S …[proved]

(ii) ∛64 × 729 = ∛64 × ∛729

∴ L.H.S, ∛64 × 729

∛46656

= 36

R.H.S, ∛64 × ∛729

= 4 × 9

= 36

L.H.S = R.H.S …[proved]

(iii) ∛- 125 × 216 = ∛- 125 × ∛216

L.H.S, ∛- 125 × 216

= ∛- 27000

= – 30

R.H.S ∛- 125 × ∛216

= – 5 × 6

= – 30

L.H.S = R.H.S …[Proved]

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

∴ L.H.S, ∛- 125 × – 1000

= ∛125000

= 50

R.H.S, ∛- 125 × ∛- 1000

= – 5 × – 10

= 50

L.H.S = R.H.S …[Proved]

Question no – (3)

Solution :

(i) 8 × 125

= ∛8 × 125

= ∛8 × ∛125

= 2 × 5

= 10

Hence, the cube root of 8 × 125 is 10.

(ii)  -1728 × 216

= ∛-1728 × ∛216

= 12 × 6

= -72

Thus, the cube root of -1728 × 216 is -72.

(iii) ∛- 27 × 2744

= ∛- 27 × 2744

= ∛2744

= – 3 × 14

= – 42

Therefore, the cube root of ∛- 27 × 2744 is -42

(iv) -729 × -15652

= ∛-729 × ∛-15652

= -9 × -25

= 225

Hence, the cube root of -729 × -15652 is 225.

Question no – (4)

Solution :

(i) We have , ∛4³ × 6³

= ∛ 2 × 2 × 2 × 2 × 3 × 2 × 3 × 2 × 3

= 2 × 2 × 2 × 3

= 24

(ii) ∛8 × 17 × 17 × 17

= ∛2 ×2 × 2 × 17 × 17

= 2 × 17

= 34

(iii)  ∛700 × 2 × 49 × 5

= ∛7 × 2 × 5 × 2 × 7 × 7 × 5 × 2 × 5

= 7 × 2 × 5

= 70

(iv) 125∛a6 – ∛125a6

= 125a2 – 5a2

= 120a2

Question no – (5)

Solution :

(i) -125/729

= -53/93

= -5/9

∴ The cube root of -125/729 is -5/9

(ii) 10648/12167

= (22)3/(23)3

= 22/23

∴ The cube root of 10648/12167 is 22/23.

(iii) -19683/24389

= (-27)3/(29)3

= 27/29

∴ The cube root of -19683/24389 is 27/29.

(iv) 686/-3456

= (-7)3/(2)3

= 7/12

∴ The cube root of 686/-3456 is 7/12.

(v) -39304/-42875

= (34)3/(35)3

= 34/35

∴ The cube root of -39304/-42875 is 34/35.

Question no – (6)

Solution :

(i) 0.001728

= ∛1728/1000000

= ∛12 × 12 × 12/100 × 100 × 100

= 12/100

= 0.12

∴ The cube root of 0.001728 is 0.12.

(ii) 0.003375

= ∛3375/1000000

= ∛15 × 15 × 15/ 100 × 100 × 100

= 15/10

= 0.15

∴ The cube root of 0.003375 is 0.15.

(iii) 0.001

= ∛1/1000 ∛1 × 1 × 1/10 × 10 × 10

= 1/10

= 0.1

∴ The cube root of 0.001 is 0.1.

(iv) 1.331

= ∛1331/1000

= ∛11 × 11 × 11/ 10 × 10 × 10

= 11/10

= 1.1

∴ The cube root of 1.331 is 1.1.

Question no – (7)

Solution :

(i) ∛27 + ∛ 0.008 + ∛0.064

= 3 + 2/10 + 4/10

= 3 + 0.2 + 0.4

= 3.6

(ii) ∛1000 + ∛0.008 – ∛0.125

= 10 + 2/10 – 5/10

= 10 + 0.2 – 0.5

= 9.7

(iii) We have, ∛0.027/0.008 ÷ √0.09/0.04 – 1

= 3/10/2/10 ÷ 3/10/2/10 – 1

= (3/10 × 10/2) ÷ (3/10 × 10/2) – 1

= 3/2 ÷ 3/2 – 1

= 3/2 × 2/3 – 1

= 1 – 1

= 0

(iv) ∛0.1 × 0.1 × 0.1 × 13 × 13

= 0.1 × 13

= 1.3

Question no – (8)

Solution :

(i) ∛729/∛1000 = ∛729/1000

L.H.S, ∛729/∛1000

= 9/10

= 0.9

R.H.S, ∛729/1000

= 9/10

= 0.9

L.H.S = R.H.S …[Proved]

(ii) ∛-512/∛343 = ∛-512/343

L.H.S = ∛512/∛343

= 8/7

R.H.S = ∛512/343

= 8/7

L.H.S = R.H.S …[Proved]

Question no – (9)

Solution :

(i) ∛125 × 27 = 3 × 5

Explanation :

∛125 × 27 = 3 × ?

∛125 × 27 = 3 × x

or, 5 × 3 = 3 × 5

Hence, ∛125 × 27 = 3 × 5

(iii) ∛1728 = 4 × 3

Explanation :

∛1728 = 4 × ?

∛1728 = 4 × x

or, 12 = 4 × 3

Thus, ∛1728 = 4 × 3

(iv) ∛480 = ∛3 × 2 × ∛20

Explanation :

∛480 = ∛3 × 2 × ∛ ?

∛480 = ∛3 × 2 × ∛x

or, ∛2 × 2 × 2 × 2 × 3 × 2 × 5 = ∛3 × 2 × 3 ∛x

or, ∛3 × 2 × 3 ∛5 × 4 = ∛3 × 2 × ∛20

Hence, ∛480 = ∛3 × 2 × ∛20

(v) ∛56 = ∛7 × ∛8

Explanation :

∛? = ∛7 × ∛8

∛x = ∛7 × ∛8

or, ∛x = √56

(vi) ∛120 = ∛4 × ∛5 × ∛6

Explanation :

∛ ? = ∛4 × ∛5 × ∛6

∛x = ∛4 × ∛5 × ∛6

or, ∛x = 3∛120

(vii) ∛27/125 = 3/5

Explanation :

∛27/125 = ?/5

∛27/125 = x/5

or, x/8 = 3/5 = x = 3

∛27/125 = 3/5

(viii) ∛729/1331 = 9/11

Explanation :

∛729/1331 = 9/?

∛729/1331 = 9/x

or, 9/11 = 9/x = x = 11

∛729/1331 = 9/11

(ix) ∛512/2197 = 8/13

Explanation :

∛512/? = 8/13

∛512/x = 8/13

or, ∛512/2197 = 8/13

x = 2197

Question no – (10)

Solution :

According to the given question,

Volume of box = 474.552 m³

One side length of box,

= ∛474.552

= ∛474552/1000

=∛78 × 78 × 78/1000

=78/10

= 7.8 m

Therefore, the length of each side of the box will be 7.8 m.

Question no – (11)

Solution :

Let, common factor = x

According to question,

(2x) ³ + (3x) ³ + (4x) ³ = 0.334125 = 334125/1000000

or, 8x³ + 27x³ + 64x³ = 334125

or, 99x³ = 334125/10000000

or, x³ = 334125/1000000 × 9 = 3375/10000000

or, x = ∛3375/100000

= ∛15 × 15 × 15/100 × 100 × 100

= 5/100

= 0.15

2x = 0.15 × 2 = 3m

3x = 0.5 × 3 = 45 m

4x = 0.15 × 4 = 6 m

Question no – (12)

Solution :

Given, Volume = 24389/216 m³

One side,

= ∛24389/216

= ∛29 × 29 × 29/6× 6 × 6

= 29/6 m

Hence, the side of a cube will be 29/6 m.

Question no – (13)

Solution :

(i) ∛36 × ∛384

= ∛3 × 3 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

= 2 × 2 × 2 × 3

= 24

(ii) ∛96 × ∛144

= ∛2 × 2 × 2 × 2 × 3 × 3 × 3 × 2 × 2 × 2 × 2

= 2 × 2 × 2 × 3

= 24

(iii) ∛100 × ∛270

= ∛10 × 10 × 10 × 3 × 3 × 3

= 10 × 3

= 30

(iv) ∛121 × ∛297

= ∛35937

= 33

Question no – (14)

Solution :

(i) 3049625 = 3375 × 729

L.H.S, ∛3049625

= 135

R.H.S, ∛3375 × 729

∛304825

= 135

L.H.S = R.H.S

(ii) 20346417 = 9261 × 2197

L.H.S, 20346417

∛20346417

= 273

R.H.S, 9261 × 2197

= ∛9261 × 2197

= √20346417

= 273

L.H.S = R.H.S

(iii) 210644875 = 42875 × 4913

L.H.S, ∛210644875

= 595

R.H.S, ∛42875 × 4913

= ∛210644875

= 595

L.H.S = R.H.S

(iv) 57066625 = 166375 × 343

L.H.S, ∛57066625

= 385

R.H.S, ∛166375 × 343

= ∛57066625

= 385

L.H.S = R.H.S

Question no – (15)

Solution :

(i) 226981

Here, Unit digit = 1

(ii) 13824

Here, Unit digit = 4

(iii) 571787

Here, Unit digit = 7

(iv) 175616

Here, Unit digit = 6

Question no – (16)

Solution :

(i) 226981

Here, 10th digit = 6

(ii) 13824

Here, 10th digit = 2

(iii) 571787

Here, 10th digit = 8

(iv) 175616

Here, 10th digit = 5

Cubes and Cube Roots Exercise 4.5 Solution :

Question no – (1)

Solution :

Given number, 7

∛7 = 1.913

So, the cube roots of 7 is 1.913.

Question no – (2)

Solution :

Given number, 70

∛70 = 4.121

Hence, the cube roots of 70 is 4.121

Question no – (3)

Solution :

Given number, 700

∛700 = 8.879

Thus, the cube root of 700 is 8.879

Question no – (4)

Solution :

Given number, 7000

∴ ∛7000 = 19.13

So, the cube root of 7000 is 19.13.

Question no – (6)

Solution :

Given number, 780

= ∛780

= 9.205

Thus, the cube root of 780 is 9.205

Question no – (7)

Solution :

Given number, 7800

= ∛7800

= 19.83

Hence, the cube root of 7800 is 19.83.

Question no – (9)

Solution :

Given number, ∛250

= ∛50 × 5

= 3.648 × 1.710

= 6.3

Hence, the cube root of 250 is 6.3.

Question no – (11)

Solution :

Given number, 9800

= ∛9800

= 21.40

So, the cube root of 9800 is 21.40.

Question no – (23)

Solution :

As per the given question,

Volume = 27 cm³

One side,

= ∛275 = ∛5 c 5 × 11 = ∛5 × ∛5 × ∛11

= 1.710 × 1.710 × 2.224

= 6.5031 m

Therefore, the length of the side of the cube will be 6.5031 m.

Next Chapter Solution :

Updated: June 13, 2023 — 1:54 pm