# Class 8 RD Sharma Solutions- Chapter 4 Cubes and Cube Roots – Exercise 4.3

**Question 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, …

### (i) 64

### (ii) 512

### (iii) 1728

**Solution:**

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free classeswhich will definitely help them in making a wise career choice in the future.Performing successive subtraction:

64 – 1 = 63

63 – 7 = 56

56 – 19 =37

37 – 37 = 0

Since subtraction is performed 4 times.

Therefore, the cube root of 64 is 4.

(ii)512Performing successive subtraction:

512 – 1 = 511

511 – 7 = 504

504 – 19 = 485

485 – 37 = 448

448 – 61 = 387

387 – 91 = 296

296 – 127 = 169

169 – 169 = 0

Since subtraction is performed 8 times.

Therefore, the cube root of 512 is 8.

(iii)1728Performing successive subtraction:

1728 – 1 = 1727

1727 – 7 = 1720

1720 – 19 = 1701

1701 – 37 = 1664

1664 – 91 = 1512

1512 – 127 = 1385

1385 – 169 = 1216

1216 – 217 = 999

999 – 271 = 728

728 – 331 = 397

397 – 397 = 0

Since subtraction is performed 12 times.

Therefore, the cube root of 1728 is 12.

**Question 2:** Using the method of successive subtraction examine whether the following numbers are perfect cubes:

### (i) 130

### (ii) 345

### (iii) 792

### (iv) 1331

**Solution:**

(i)130Performing successive subtraction:

130 – 1 = 129

129 – 7 = 122

122 – 19 = 103

103 – 37 = 66

66 – 61 = 5

Since the next number to be subtracted is 91, which is greater than 5

Therefore,130 is not a perfect cube.

(ii)345Performing successive subtraction:

345 – 1 = 344

344 – 7 = 337

337 – 19 = 318

318 – 37 = 281

281 – 61 = 220

220 – 91 = 129

129 – 127 = 2

Since the next number to be subtracted is 169, which is greater than 2

Therefore, 345 is not a perfect cube

(iii) 792Performing successive subtraction:

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

Since the next number to be subtracted is 271, which is greater than 63

Therefore, 792 is not a perfect cube

(iv)1331Performing successive subtraction:

1331 – 1 = 1330

1330 – 7 = 1323

1323 – 19 = 1304

1304 – 37 = 1267

1267 – 61 = 1206

1206 – 91 = 1115

1115 – 127 = 988

988 – 169 = 819

819 – 217 = 602

602 – 271 = 331

331 – 331 = 0

Since subtraction is performed 11 times,

Therefore, Cube root of 1331 is 11

Hence, 1331 is a perfect cube.

**Question 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 the previous question, there are three numbers that are not perfect cubes.

(i)130Performing successive subtraction:

130 – 1 = 129

129 – 7 = 122

122 – 19 = 103

103 – 37 = 66

66 – 61 = 5

The next number which is to be subtracted is 91, which is greater than 5

Since, 130 is not a perfect cube.

Therefore, to make it a perfect cube we have to subtract 5.

130 – 5 = 125

125 is a perfect cube of 5.

(ii)345Performing successive subtraction:

345 – 1 = 344

344 – 7 = 337

337 – 19 = 318

318 – 37 = 281

281 – 61 = 220

220 – 91 = 129

129 – 127 = 2

The next number which is to be subtracted is 169, which is greater than 2

Since, 345 is not a perfect cube.

Therefore, to make it a perfect cube we have to subtract 2.

345 – 2 = 343

343 is a perfect cube of 7.

(iii)792Performing successive subtraction:

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

The next number which is to be subtracted is 271, which is greater than 63

Since, 792 is not a perfect cube.

Therefore, to make it a perfect cube we have to subtract 63.

792 – 63 = 729

729 is a perfect cube of 9.

**Question 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)343By prime factorizing 343, we get

∛343 = ∛ (7 × 7 × 7) = 7

Therefore, the cube root of 343 is 7

(ii)2744By prime factorizing 2744, we get

∛2744 = ∛ (2 × 2 × 2 × 7 × 7 × 7)

∛2744 = ∛ (23 × 73) = 2 × 7 = 14

Therefore, the cube root of 2744 is 14

(iii)4913By prime factorizing 4913, we get

∛4913 = ∛ (17 × 17 × 17) = 17

Therefore, the cube root of 4913 is 17

(iv) 1728By prime factorizing 1728, we get

∛1728 = ∛(2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3)

∛1728 = ∛ (2

^{3 }× 2^{3 }× 3^{3}) = 2 × 2 × 3 = 12Therefore, the cube root of 1728 is 12

(v)35937By prime factorizing 35937, we get

∛35937 = ∛ (3 × 3 × 3 × 11 × 11 × 11)

∛35937 = ∛ (3

^{3 }× 11^{3}) = 3 × 11 = 33Therefore, the cube root of 35937 is 33

(vi) 17576By prime factorizing 17576, we get

∛17576 = ∛ (2 × 2 × 2 × 13 × 13 × 13)

∛17576 = ∛ (2

^{3 }× 13^{3}) = 2 × 13 = 26Therefore, the cube root of 17576 is 26

(vii) 134217728By prime factorizing 134217728, we get

∛134217728 = ∛ (2

^{27}) = 2^{9}= 512Therefore, the cube root of 134217728 is 512

(viii)48228544By prime factorizing 48228544, we get

∛48228544 = ∛ (2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 × 7 × 13 × 13 × 13)

∛48228544 = ∛ (2

^{3 }× 2^{3 }× 7^{3 }× 13^{3}) = 2 × 2 × 7 × 13 = 364Therefore, the cube root of 48228544 is 364

(ix)74088000By prime factorizing 74088000, we get

∛74088000 = ∛ (2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5 × 7 × 7 × 7)

∛74088000 = ∛ (2

^{3 }× 2^{3 }× 3^{3 }× 5^{3 }× 7^{3}) = 2 × 2 × 3 × 5 × 7 = 420Therefore, the cube root of 74088000 is 420

(x)157464By prime factorizing 157464, we get

∛157464 = ∛ (2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3)

∛157464 = ∛ (2

^{3 }× 3^{3 }× 3^{3 }× 3^{3}) = 2 × 3 × 3 × 3 = 54Therefore, the cube root of 157464 is 54

(xi) 1157625By prime factorizing 1157625, we get

∛1157625 = ∛ (3 × 3 × 3 × 5 × 5 × 5 × 7 × 7 × 7)

∛1157625 = ∛ (3

^{3 }× 5^{3 }× 7^{3}) = 3 × 5 × 7 = 105Therefore, the cube root of1157625 is 105

(xii)33698267By prime factorizing 33698267, we get

∛33698267 = ∛ (17 × 17 × 17 × 19 × 19 × 19)

∛33698267 = ∛ (17

^{3 }× 19^{3}) = 17 × 19 = 323Therefore, the cube root of 33698267 is 323

**Question 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:**

By prime factorizing 3600, we get

3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5

By forming groups in triplet of equal factors we get, 3600 = (2 × 2 × 2) × (3 × 3) × (5 × 5) × 2

Since, 2, 3 and 5 cannot form a triplet of equal factors.

Therefore, 3600 must be multiplied with 60 (2 × 2 × 3 × 5) to get a perfect cube.

3600 × 60 = 216000

Cube root of 216000 is

∛216000 = ∛ (60 × 60 × 60)

∛216000 = ∛ (60

^{3}) = 60Therefore, the smallest number which when multiplied with 3600 makes a perfect cube is 60.

**Question 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:**

By prime factorizing 210125, we get

210125 = 5 × 5 × 5 × 41 × 41

By forming groups in triplet of equal factors we get, 210125 = (5 × 5 × 5) × (41 × 41)

Since, 41 cannot form a triplet of equal factors.

Therefore, 210125 must be multiplied with 41 to get a perfect cube.

210125 × 41 = 8615125

Now, finding the cube root of 8615125

By using the prime factorization method, we get

8615125 = 5 × 5 × 5 × 41 × 41 × 41

Therefore, Cube root of product = ∛8615125 = ∛ (5 × 41) = 205

**Question 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:**

By prime factorizing 8192, we get

8192 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 23 × 23 × 23 × 2

By forming groups in triplet of equal factors we get, 8192 = (2×2×2)×(2×2×2)×(2×2×2)×(2×2×2)×2

Since, 2 cannot form a triplet of equal factors.

Therefore, 8192 must be divided by 2 to get a perfect cube.

8192/2 = 4096

Now, finding the cube root of 4096

By using the prime factorization method, we get

4096 = 2×2×2×2×2×2×2×2×2×2×2×2 = 2

^{3}×2^{3}×2^{3}×2^{3}Therefore, Cube root of 4096 = ∛4096 = ∛ (2

^{3}×2^{3}×2^{3}×2^{3}) = 2×2×2×2 = 16

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

**Solution:**

Given, ratio of number is 1:2:3

Therefore, Let the number be x, 2x and 3x

According to the question, sum of their cube is 98784

x

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

^{3}+ 8x^{3}+ 27x^{3}= 9878436x

^{3}= 98784x

^{3}= 98784/36x = 2744

x = ∛2744 = ∛ (2 × 2 × 2 × 7 × 7 × 7)

x = 2×7

x = 14

So, the respected numbers are,

x = 14

2x = 2 × 14 = 28

3x = 3 × 14 = 42

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

**Solution:**

Given, the volume of cube = 9261000 m

^{3}Let the side of the cube be ‘x’ meter

Therefore, x

^{3}= 9261000Taking cube root on both the side,

x = ∛9261000 = ∛ (2×2×2×3×3×3×5×5×5×7×7×7) = ∛ (23×33×53×73) = 2×3×5×7 = 210

Hence, the side of cube is 210 meter