Download free pdf of RD Sharma Class 8 Maths from the links provided below. Students can refer to RD Sharma textbook and then practice the RD Sharma Class 8 Solutions provided by BYJUâ€™S expert team. The solutions are solved using simple techniques for easy understanding, which help in achieving high marks in their examination. In Exercise 4.3 of RD Sharma Solutions for Class 8 Maths Chapter 4 Cubes and Cube Roots, we shall discuss problems based on cube roots and cube root of a natural number using units digit method, cube root of a perfect cube by factors.

## Download PDF of RD Sharma Solutions for Class 8 Maths Exercise 4.3 Chapter 4 Cubes and Cube Roots

### Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 4.3 Chapter 4 Cubes and Cube Roots

**1. Find the cube roots of the following numbers by successive subtraction of numbers:
1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, â€¦
(i) 64
(ii) 512
(iii) 1728**

**Solution:**

**(i)** 64

Letâ€™s perform subtraction

64 â€“ 1 = 63

63 â€“ 7 = 56

56 â€“ 19 =37

37 â€“ 37 = 0

Subtraction is performed 4 times.

âˆ´ Cube root of 64 is 4.

**(ii)** 512

Letâ€™s perform subtraction

512 â€“ 1 = 511

511 â€“ 7 = 504

504 â€“ 19 = 485

485 â€“ 37 = 448

448 â€“ 61 = 387

387 â€“ 91 = 296

296 â€“ 127 = 169

169 â€“ 169 = 0

Subtraction is performed 8 times.

âˆ´ Cube root of 512 is 8.

**(iii)** 1728

Letâ€™s perform subtraction

1728 â€“ 1 = 1727

1727 â€“ 7 = 1720

1720 â€“ 19 = 1701

1701 â€“ 37 = 1664

1664 â€“ 91 = 1512

1512 â€“ 127 = 1385

1385 â€“ 169 = 1216

1216 â€“ 217 = 999

999 â€“ 271 = 728

728 â€“ 331 = 397

397 â€“ 397 = 0

Subtraction is performed 12 times.

âˆ´ Cube root of 1728 is 12.

**2. Using the method of successive subtraction examine whether or not the following numbers are perfect cubes:
(i) 130
(ii) 345
(iii) 792
(iv) 1331**

**Solution:**

**(i)** 130

Letâ€™s perform subtraction

130 â€“ 1 = 129

129 â€“ 7 = 122

122 â€“ 19 = 103

103 â€“ 37 = 66

66 â€“ 61 = 5

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

âˆ´130 is not a perfect cube.

**(ii)** 345

Letâ€™s perform subtraction

345 â€“ 1 = 344

344 â€“ 7 = 337

337 â€“ 19 = 318

318 â€“ 37 = 281

281 â€“ 61 = 220

220 â€“ 91 = 129

129 â€“ 127 = 2

Next number to be subtracted is 169, which is greater than 2

âˆ´ 345 is not a perfect cube

**(iii)** 792

Letâ€™s perform subtraction

792 â€“ 1 = 791

791 â€“ 7 = 784

784 â€“ 19 = 765

765 â€“ 37 = 728

728 â€“ 61 = 667

667 â€“ 91 = 576

576 â€“ 127 = 449

449 â€“ 169 = 280

280 â€“ 217 = 63

Next number to be subtracted is 271, which is greater than 63

âˆ´ 792 is not a perfect cube

**(iv)** 1331

Letâ€™s perform subtraction

1331 â€“ 1 = 1330

1330 â€“ 7 = 1323

1323 â€“ 19 = 1304

1304 â€“ 37 = 1267

1267 â€“ 61 = 1206

1206 â€“ 91 = 1115

1115 â€“ 127 = 988

988 â€“ 169 = 819

819 â€“ 217 = 602

602 â€“ 271 = 331

331 â€“ 331 = 0

Subtraction is performed 11 times

Cube root of 1331 is 11

âˆ´ 1331 is a perfect cube.

**3. Find the smallest number that must be subtracted from those of the numbers in question 2 which are not perfect cubes, to make them perfect cubes. What are the corresponding cube roots?**

**Solution:**

In previous question there are three numbers which are not perfect cubes.

**(i)** 130

Letâ€™s perform subtraction

130 â€“ 1 = 129

129 â€“ 7 = 122

122 â€“ 19 = 103

103 â€“ 37 = 66

66 â€“ 61 = 5

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

Since, 130 is not a perfect cube. So, to make it perfect cube we subtract 5 from the given number.

130 â€“ 5 = 125 (which is a perfect cube of 5)

**(ii)** 345

Letâ€™s perform subtraction

345 â€“ 1 = 344

344 â€“ 7 = 337

337 â€“ 19 = 318

318 â€“ 37 = 281

281 â€“ 61 = 220

220 â€“ 91 = 129

129 â€“ 127 = 2

Next number to be subtracted is 169, which is greater than 2

Since, 345 is not a perfect cube. So, to make it a perfect cube we subtract 2 from the given number.

345 â€“ 2 = 343 (which is a perfect cube of 7)

**(iii)** 792

Letâ€™s perform subtraction

792 â€“ 1 = 791

791 â€“ 7 = 784

784 â€“ 19 = 765

765 â€“ 37 = 728

728 â€“ 61 = 667

667 â€“ 91 = 576

576 â€“ 127 = 449

449 â€“ 169 = 280

280 â€“ 217 = 63

Next number to be subtracted is 271, which is greater than 63

Since, 792 is not a perfect cube. So, to make it a perfect cube we subtract 63 from the given number.

792 â€“ 63 = 729 (which is a perfect cube of 9)

**4. Find the cube root of each of the following natural numbers:
(i) 343 (ii) 2744**

**(iii) 4913 (iv) 1728
(v) 35937 (vi) 17576
(vii) 134217728 (viii) 48228544
(ix) 74088000 (x) 157464
(xi) 1157625 (xii) 33698267**

**Solution:**

**(i)** 343

By using prime factorization method

âˆ›343 = âˆ› (7Ã—7Ã—7) = 7

**(ii)** 2744

By using prime factorization method

âˆ›2744 = âˆ› (2Ã—2Ã—2Ã—7Ã—7Ã—7) = âˆ› (2^{3}Ã—7^{3}) = 2Ã—7 = 14

**(iii)** 4913

By using prime factorization method,

âˆ›4913 = âˆ› (17Ã—17Ã—17) = 17

**(iv)** 1728

By using prime factorization method,

âˆ›1728 = âˆ›(2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—3Ã—3Ã—3) = âˆ› (2^{3}Ã—2^{3}Ã—3^{3}) = 2Ã—2Ã—3 = 12

**(v)** 35937

By using prime factorization method,

âˆ›35937 = âˆ› (3Ã—3Ã—3Ã—11Ã—11Ã—11) = âˆ› (3^{3}Ã—11^{3}) = 3Ã—11 = 33

**(vi)** 17576

By using prime factorization method,

âˆ›17576 = âˆ› (2Ã—2Ã—2Ã—13Ã—13Ã—13) = âˆ› (2^{3}Ã—13^{3}) = 2Ã—13 = 26

**(vii)** 134217728

By using prime factorization method

âˆ›134217728 = âˆ› (2^{27}) = 2^{9} = 512

**(viii)** 48228544

By using prime factorization method

âˆ›48228544 = âˆ› (2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—7Ã—7Ã—7Ã—13Ã—13Ã—13) = âˆ› (2^{3}Ã—2^{3}Ã—7^{3}Ã—13^{3}) = 2Ã—2Ã—7Ã—13 = 364

**(ix)** 74088000

By using prime factorization method

âˆ›74088000 = âˆ› (2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—3Ã—3Ã—3Ã—5Ã—5Ã—5Ã—7Ã—7Ã—7) = âˆ› (2^{3}Ã—2^{3}Ã—3^{3}Ã—5^{3}Ã—7^{3}) = 2Ã—2Ã—3Ã—5Ã—7 = 420

**(x)** 157464

By using prime factorization method

âˆ›157464 = âˆ› (2Ã—2Ã—2Ã—3Ã—3Ã—3Ã—3Ã—3Ã—3Ã—3Ã—3Ã—3) = âˆ› (2^{3}Ã—3^{3}Ã—3^{3}Ã—3^{3}) = 2Ã—3Ã—3Ã—3 = 54

**(xi)** 1157625

By using prime factorization method

âˆ›1157625 = âˆ› (3Ã—3Ã—3Ã—5Ã—5Ã—5Ã—7Ã—7Ã—7) = âˆ› (3^{3}Ã—5^{3}Ã—7^{3}) = 3Ã—5Ã—7 = 105

**(xii)** 33698267

By using prime factorization method

âˆ›33698267 = âˆ› (17Ã—17Ã—17Ã—19Ã—19Ã—19) = âˆ› (17^{3}Ã—19^{3}) = 17Ã—19 = 323

**5. Find the smallest number which when multiplied with 3600 will make the product a perfect cube. Further, find the cube root of the product.**

**Solution:**

Firstly letâ€™s find the prime factors for 3600

3600 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 5 Ã— 5

= 2^{3}Â Ã— 3^{2}Â Ã— 5^{2}Â Ã— 2

Since only one triples is formed and three factors remained ungrouped in triples.

The given number 3600 is not a perfect cube.

To make it a perfect cube we have to multiply it by (2 Ã— 2 Ã— 3 Ã— 5) = 60

3600 Ã— 60 = 216000

Cube root of 216000 is

âˆ›216000 = âˆ› (60Ã—60Ã—60) = âˆ› (60^{3}) = 60

âˆ´ the smallest number which when multiplied with 3600 will make the product a perfect cube is 60 and the cube root of the product is 60.

**6. Multiply 210125 by the smallest number so that the product is a perfect cube. Also, find out the cube root of the product.**

**Solution:**

The prime factors of 210125 are

210125 = 5 Ã— 5 Ã— 5 Ã— 41 Ã— 41

Since, one triples remained incomplete, 210125 is not a perfect cube.

To make it a perfect cube we need to multiply the factors by 41, we will get 2 triples as 23Â and 41^{3}.

And the product become:

210125 Ã— 41 = 8615125

8615125 = 5 Ã— 5 Ã— 5 Ã— 41 Ã— 41 Ã— 41

Cube root of product = âˆ›8615125 = âˆ› (5Ã—41) = 205

**7. What is the smallest number by which 8192 must be divided so that quotient is a perfect cube? Also, find the cube root of the quotient so obtained.**

**Solution**:

The prime factors of 8192 are

8192 = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2 = 2^{3}Ã—2^{3}Ã—2^{3}Ã—2

Since, one triples remain incomplete, hence 8192 is not a perfect cube.

So, we divide 8192 by 2 to make its quotient a perfect cube.

8192/2 = 4096

4096 = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2 = 2^{3}Ã—2^{3}Ã—2^{3}Ã—2^{3}

Cube root of 4096 = âˆ›4096 = âˆ› (2^{3}Ã—2^{3}Ã—2^{3}Ã—2^{3}) = 2Ã—2Ã—2Ã—2 = 16

**8. Three numbers are in the ratio 1:2:3. The sum of their cubes is 98784. Find the numbers.**

**Solution:**

Let us consider the ratio 1:2:3 as x, 2x and 3x

According to the question,

X^{3}Â + (2x)^{ 3}Â + (3x)^{ 3}Â = 98784

x^{3}Â + 8x^{3}Â + 27x^{3}Â = 98784

36x^{3}Â = 98784

x^{3} = 98784/36

= 2744

x = âˆ›2744 = âˆ› (2Ã—2Ã—2Ã—7Ã—7Ã—7) = 2Ã—7 = 14

So, the numbers are,

x = 14

2x = 2Â Ã— 14 = 28

3x = 3Â Ã— 14 = 42

9. **The volume of a cube is 9261000 m ^{3}. Find the side of the cube.**

Given, volume of cube = 9261000 m^{3}

Let us consider the side of cube be â€˜aâ€™ metre

So, a^{3} = 9261000

a = âˆ›9261000 = âˆ› (2Ã—2Ã—2Ã—3Ã—3Ã—3Ã—5Ã—5Ã—5Ã—7Ã—7Ã—7) = âˆ› (2^{3}Ã—3^{3}Ã—5^{3}Ã—7^{3}) = 2Ã—3Ã—5Ã—7 = 210

âˆ´ the side of cube = 210 metre