## RD Sharma Solutions Class 8 Chapter 4 Exercise 4.3

Q1. 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

Answer:

(i) We have:

64

__Â 1Â __

63

__Â 7__

56

__19__

37

__37__

0

Subtraction is performed 4 times.

\(\sqrt[3]{64}\)

(ii) We have:

512

__Â Â 1Â __

511

__Â Â Â 7__

504

__Â Â 19__

485

__Â 37__

448

__Â Â 61__

387

__Â Â 91__

296

__127__

169

__169__

0

Subtraction is performed 8 times.

\(\sqrt[3]{512}\)

(iii) We have:

1728

__Â Â Â Â Â Â 1Â __

1727

__Â Â Â Â Â Â 7__

1720

__Â Â Â Â 19__

1701

__Â Â 37__

1664

__Â Â Â 61__

1603

__Â Â Â 91__

1512

__Â 127__

1385

__Â 169__

1216

__Â 217__

999

__Â 271__

728

__331__

397

__397__

0

Subtraction is performed 12 times.

\(\sqrt[3]{1728}\)

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Q2. Using the method of successive subtraction examine whether or not the following numbers are perfect cubes:

(i) 130 (ii) 345 (iii) 792 (iv) 1331

Answer:

(i)

We have:

130

__Â Â Â 1Â __

129

__Â Â Â 7__

122

__Â 19__

103

__Â 37__

66

__61__

5

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

Hence, 130 is not a perfect cube.

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(ii) We have:

345

__Â Â Â 1Â __

344

__Â Â 7__

337

__Â Â 19__

318

__Â Â 37__

281

__Â Â 61__

220

__Â Â 91__

129

__127__

2

Therefore, the next number to be subtracted is 161, which is greater than 2.

Hence, 345 is not a perfect cube

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(iii) We have:

792

__Â Â Â 1Â __

791

__Â Â Â 7__

784

__Â Â 19__

765

__Â 37__

728

__Â 61__

667

__Â Â 91__

576

__127__

449

__169__

280

__217__

63

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

Hence, 792 is not a perfect cube

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(iv) We have:

1331

__Â Â Â Â Â Â 1Â __

1330

__Â Â Â Â Â Â 7__

1323

__Â Â Â Â 19__

1304

__Â Â Â Â 37__

1267

__Â Â Â 61__

1206

__Â Â Â 91__

1115

__Â 127__

988

__169__

819

__217__

602

__271__

331

__331__

0

The subtraction is performed 11 times.

Therefore, \(\sqrt[3]{1331}\)

Thus, 1331 is a perfect cube.

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Q3. 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?

Answer :

(i)

We have:

130

__Â Â Â Â Â Â 1Â __

129

__Â Â Â Â Â Â 7__

122

__Â Â Â 19__

103

__Â Â Â 37__

66

__Â Â Â 61__

5

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

130 is not a perfect cube.

However, if we subtract 5 from 130, we will get 0 on performing successive subtraction and the number will become a perfect cube.

If we subtract 5 from 125, we get 125. Now, find the cube root using successive subtraction.

We have:

125

__Â Â Â 1Â __

125

__Â Â Â 7__

117

__Â Â Â 19__

98

__Â 37__

61

__Â 61__

0

The subtraction is performed 5 times.

\(\sqrt[3]{125}\)

Thus, it is a perfect cube.

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(ii) We have:

345

__Â Â Â 1__

344

__Â Â Â Â 7Â __

337

__Â 19 __

318

__Â 37 __

281

__Â 61 __

220

__Â 91 __

129

__127__

2

Since, the next number to be subtracted is 161, which is greater than 2.

Thus, 345 is not a perfect cube.

However, if we subtract 2 from 345, we will get 0 on performing successive subtraction and the number will become a perfect cube.

If we subtract 2 from 345, we get 343. Now, find the cube root using successive subtraction.

343

__Â Â Â 1__

342

__Â Â Â 7__

335

__Â Â 19__

316

__Â Â 37__

279

__Â Â 61__

218

__Â Â 91__

127

__127__

0

The subtraction is performed 7 times.

\(\sqrt[3]{343}\)

Thus, it is a perfect cube.

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(iii) We have:

792

__Â Â Â 1__

791

__Â Â Â 7__

784

__Â 19__

765

__Â Â 37__

728

__Â Â 61__

667

__Â Â 91__

576

__Â 127 __

449

__169__

280

__Â 217 __

63

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

792 is not a perfect cube.

However, if we subtract 63 from 792, we will get 0 on performing successive subtraction and the number will become a perfect cube.

If we subtract 63 from 792, we get 729.

Now, find the cube root using the successive subtraction. We have:

729

__Â Â Â 1__

728

__Â Â Â 7__

721

__Â Â 19__

702

__Â Â 37__

665

__Â Â 61__

604

__Â Â 91__

513

__127__

386

__169__

217

__217__

0

The subtraction is performed 9 times.

\(\sqrt[3]{729}\)

Thus, it is a perfect cube.

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Q4. Find the cube root of each of the following natural numbers:

(i) 343

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(ii) 2744

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(iii) 4913

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(iv) 1728

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(v) 35937

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(vi) 17576

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(vii) 134217728

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(viii) 48228544

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(ix) 74088000

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(x) 157464

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(xi) 1157625

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(xii) 33698267

Answer :

(i) Cube root using units digit:

Let us consider 343.

The unit digit is 3; therefore, the unit digit in the cube root of 343 is 7.

There is no number left after striking out the units, tens and hundreds digits of the given number; therefore, the cube root of 343 is 7.

Hence, \(\sqrt[3]{343}\)

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(ii) Cube root using units digit:

Let us consider 2744.

The unit digit is 4; therefore, the unit digit in the cube root of 2744 is 4.

After striking out the units, tens and hundreds digits of the given number, we are left with 2.

Now, 1 is the largest number whose cube is less than or equal to 2.

Therefore, the tens digit of the cube root of 2744 is 1.

Hence, \(\sqrt[3]{2744}\)

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(iii) Cube root using units digit:

Let us consider 4913.

The unit digit is 3; therefore, the unit digit in the cube root of 4913 is 7.

After striking out the units, tens and hundreds digits of the given number, we are left with 4.

Now, 1 is the largest number whose cube is less than or equal to 4.

Therefore, the tens digit of the cube root of 4913 is 1.

Hence, \(\sqrt[3]{4913}\)

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(iv) Cube root using units digit:

Let us consider 1728.

The unit digit is 8; therefore, the unit digit in the cube root of 1728 is 2.

After striking out the units, tens and hundreds digits of the given number, we are left with 1.

Now, 1 is the largest number whose cube is less than or equal to 1.

Therefore, the tens digit of the cube root of 1728 is 1.

Hence, \(\sqrt[3]{1728}\)

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(v) Cube root using units digit:

Let us consider 35937.

The unit digit is 7; therefore, the unit digit in the cube root of 35937 is 3.

After striking out the units, tens and hundreds digits of the given number, we are left with 35.

Now, 3 is the largest number whose cube is less than or equal to 35 ( 33 < 35 < 43).

Therefore, the tens digit of the cube root of 35937 is 3.

Hence, \(\sqrt[3]{35937}\)

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(vi) Cube root using units digit:

Let us consider the number 17576.

The unit digit is 6; therefore, the unit digit in the cube root of 17576 is 6.

After striking out the units, tens and hundreds digits of the given number, we are left with 17.

Now, 2 is the largest number whose cube is less than or equal to 17 (23 < 17 < 33).

Therefore, the tens digit of the cube root of 17576 is 2. Hence,’ \(\sqrt[3]{17576}\)

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(vii) Cube root by factors:

On factorising 134217728 into prime factors, we get:

134217728=2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

On grouping the factors in triples of equal factors, we get: 134217728 = {2 x 2 x 2} x {2 x 2 x 2} x {2 x 2 x 2} x {2 x 2 x 2} x {2 x 2 x 2} x {2 x 2 x 2} x{2 x 2 x 2} x{2 x 2 x 2} x{2 x 2 x 2}

Now, taking one factor from each triple, we get: \(\sqrt[3]{134217728}\)

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(viii) Cube root by factors:

On factorising 48228544 into prime factors, we get:

48228544 = 2 x 2 x 2 x 2 x 2 x 2 x 7 x 7 x 7 x 13 x 13 x 13

On grouping the factors in triples of equal factors, we get:

48228544 = {2 x 2 x 2} x {2 x 2 x 2} x {7 x 7 x 7} x {13 x 13 x 13}

Now, taking one factor from each triple, we get: \(\sqrt[3]{ 48228544}\)

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(ix) Cube root by factors:

On factorising 74088000 into prime factors, we get: 74088000 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 5 x 5 x 5 x 7 x 7 x 7

On grouping the factors in triples of equal factors, we get: 74088000 = {2 x 2 x 2} x {2 x 2 x 2} x {3 x 3 x 3} x {5 x 5 x 5} x {7 x 7 x 7}

Now, taking one factor from each triple, we get: \(\sqrt[3]{ 74088000}\)

(x) Cube root using units digit:

Let is consider 157464.

The unit digit is 4; therefore, the unit digit in the cube root of 157464 is 4.

After striking out the units, tens and hundreds digits of the given number, we are left with 157.

Now, 5 is the largest number whose cube is less than or equal to 157 (53 < 157 < 63).

Therefore, the tens digit of the cube root 157464 is 5. Hence, \(\sqrt[3]{ 157464}\)

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(xi) Cube root by factors:

On factorising 1157625 into prime factors, we get: 1157625 = 3 x 3 x 3 x 5 x 5 x 5 x 7 x 7 x 7

On grouping the factors in triples of equal factors, we get:

1157625 = {3 x 3 x 3} x {5 x 5 x 5} x {7 x 7 x 7}

Now, taking one factor from each triple, we get:

\(\sqrt[3]{1157625}\)

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(xii) Cube root by factors:

On factorising 33698267 into prime factors, we get: 33698267 = 17 x 17 x 17 x 19 x 19 x 19

On grouping the factors in triples of equal factors, we get: 33698267 = {17 x 17 x 17} x {19 x 19 x 19}

Now, taking one factor from each triple, we get:

\(\sqrt[3]{ 33698267}\)

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Q5. Find the smallest number which when multiplied with 3600 will make the product a perfect cube. Further, find the cube root of the product.

Answer:

On factorising 3600 into prime factors, we get:

3600 = 2 x 2 x 2 x 2 x 3 x 3 x 5 x 5

On grouping the factors in triples of equal factors, we get:

3600 = {2 x 2 x 2} x 2 x 3 x 3 x 5 x 5

It is evident that the prime factors of 3600 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 3600 is not a perfect cube.

However, if the number is multiplied by (2 x 2 x 3 x 5 = 60), the factors can be grouped into triples of equal factors such that no factor is left over.

Hence, the number 3600 should be multiplied by 60 to make it a perfect cube.

Also, the product is given as:

3600 x 60 ={2 x 2 x 2} x 2 x 3 x 3 x 5 x 5 x 60

216000 = {2 x 2 x 2} x 2 x 3 x 3 x 5 x 5 x (2 x 2 x 3 x 5)

216000 ={2 x 2 x 2}x {2 x 2 x 2} x {3 x 3 x 3} x {5 x 5 x 5}

To get the cube root of the produce 216000, take one factor from each triple.

Cube root = 2 x 2 x 3 x 5 = 60

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Q6. Multiply 210125 by the smallest number so that the product is a perfect cube. Also, find out the cube root of the product.

Answer:

On factorising 210125 into prime factors, we get: 210125 = 5 x 5 x 5 x 41 x 41

On grouping the factors in triples of equal factors, we get: 210125 = {5 x 5 x x 41 x 41

It is evident that the prime factors of 210125 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 210125 is not a perfect cube. However, if the number is multiplied by 41, the factors can be grouped into triples of equal factors such that no factor is left over.

Hence, the number 210125 should be multiplied by 41 to make it a perfect cube. Also, the product is given as: 210125 x 41 = {5 x 5 x 5} x {41 x 41 x 41} 8615125 = {5 x 5 x x {41 x 41 x 41} To get the cube root of the produce 8615125, take one factor from each triple. The cube root is 5 x 41 = 205. Hence, the required numbers are 41 and 205.

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Q7. 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.

Answer : Â Â Â Â

On factorising 8192 into prime factors, we get:

8192 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

On grouping the factors in triples of equal factors, we get:

8192 = {2 x 2 x 2} x {2 x 2 x 2} x {2 x 2 x 2} x {2 x 2 x 2} x 2

It is evident that the prime factors of 8192 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 8192 is not a perfect cube. However, if the number is divided by 2, the factors can be grouped into triples of equal factors such that no factor is left over.

Hence, the number 8192 should be divided by 2 to make it a perfect cube.

Also, the quotient is given as:

\(\frac{8192}{2} = \frac{{2 \times 2 \times 2} \times {2 \times 2 \times 2} \times {2 \times 2 \times 2} \times {2 \times 2 \times 2} \times 2}{2}\)

4096 = {2 x 2 x 2} x {2 x 2 x 2} x {2 x 2 x 2} x {2 x 2 x 2}

To get the cube root of the quotient 4096, take one factor from each triple. We get:

Cube root = 2 x 2 x 2 x 2 = 16

Hence, the required numbers are 2 and 16.

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Q8. Three numbers are in the ratio 1 : 2 : 3. The sum of their cubes is 98784. Find the numbers.

Answer:

Let the numbers be x, 2x and 3x

Therefore

\(x^{3} + (2x)^{3} + (3x)^{3} = 98784 \\ => x^{3} + 8x^{3} + 27x^{3} = 98784 \\ => 36 x^{3} = 98784 \\ => x^{3} = \frac{98784}{36} \\ => x^{3} = 2744 \\ => x = \sqrt[3]{2744} = \sqrt[3]{{2 \times 2 \times 2} \times {7 \times 7 \times 7}} = 2 \times 7 = 14 \\\)

Hence, the numbers are 14, (2 x 14 = 28) and (3 x 14 = 42)

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Q9. The volume of a cube is 9261000 m3. Find the side of the cube.

Answer:

Volume of a cube is given by:

V = s^{3}, where s = Side of the cube

It is given that the volume of the cube is 9261000 m^{3}; therefore, we have:

s^{3} = 9261000

Let us find the cube root of 9261000 using prime factorisation:

9261000 = 2 x 2 x 2 x 3 x 3 x 3 x 5 x 5 x 5 x 7 x 7 x 7 = {2 x 2 x 2} x {3 x 3 x 3} x {5 x 5 x 5} x {7 x 7 x 7}

9261000 could be written as triples of equal factors; therefore, we get: Cube root = 2 x 3 x 5 x 7 = 210

Therefore

s^{3} = 9261000

s = (9261000)^{(1/3)} = 210

Hence, the length of the side of cube is 210 m.