RD Sharma Solutions Class 8 Cubes And Cube Roots Exercise 4.2

RD Sharma Solutions Class 8 Chapter 4 Exercise 4.2

RD Sharma Class 8 Solutions Chapter 4 Ex 4.2 PDF Free Download

Q1. Find the cubes of:

(i) – 11    (ii) – 12     (iii) – 21

 Answer:

(i) Cube of – 11 is given as: (-11)3 = – 11 x – 11 x – 11 = – 1331

Thus, the cube of 11 is (-1331).

(ii) Cube of – 12 is given as: (-12) 3 = – 12 x – 12 x – 12 = – 1728

Thus, the cube of – 12 is (- 1728).

(iii) Cube of – 21 is given as:

(- 21) 3 = – 21 x – 21 x – 21 = – 9261

Thus the cube of – 21 is (- 9261).

 

Q2. Which of the following numbers are cubes of negative integers?

(i) – 64   

(ii) – 1056  

(iii) – 2197  

(iv) – 2744   

(v) – 42875

Answer:

In order to check if a negative number is a perfect cube, first check if the corresponding positive integer is a perfect cube. Also, for any positive integer m, – m3 is the cube of – m.

(i)  On factorizing 64 into prime factors, we get:

64 = 2 x 2 x 2 x 2 x 2 x 2

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

64 = {2 x 2 x 2} x {2 x 2 x 2}

It is evident that the prime factors of 64 can be grouped into triples of equal factors and no factor is left over. Therefore, 64 is a perfect cube. This implies that – 64 is also a perfect cube.

Now, collect one factor from each triplet and multiply, we get: 2 x 2=4

This implies that 64 is a cube of 4. Thus, – 64 is the cube of -4.

 

(ii) On factorising 1056 into prime factors, we get:

1056 = 2 x 2 x 2 x 2 x 2 x 3 x 11

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

1056 ={2 x 2 x 2}x 2 x 2 x 3 x 11

It is evident that the prime factors of 1056 cannot be grouped into triples of equal factors such that no factor is left over.

Therefore, 1056 is not a perfect cube. This implies that – 1056 is not a perfect cube as well.

 

(iii) On factorising 2197 into prime factors, we get:

2197 =13 x 13 x 13

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

2197 = {13 x 13 x13}

It is evident that the prime factors of 2197 can be grouped into triples of equal factors and no factor is left over. Therefore, 2197 is a perfect cube. This implies that – 2197 is also a perfect cube.

Now, collect one factor from each triplet and multiply, we get 13. This implies that 2197 is a cube of 13. Thus, -2197 is the cube of – 13.

 

(iv) On factorizing 2744 into prime factors, we get:

2744 = 2 x 2 x 2 x 7 x 7 x 7

On grouping the factors in triples of equal factors, we get: 2744 = {2 x 2 x 2} x {7 x 7 x 7}

It is evident that the prime factors of 2744 can be grouped into triples of equal factors and no factor is left over. Therefore, 2744 is a perfect cube. This implies that – 2744 is also a perfect cube.

Now, collect one factor from each triplet and multiply, we get:

2 x 7 = 14

This implies that 2744 is a cube of 14.

Thus, – 2744 is the cube of – 14.

 

(v) On factorizing 42875 into prime factors, we get:

42875 = 5 x 5 x 5 x 7 x 7 x 7

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

42875 = {5 x 5 x 5} x {7 x 7 x 7}

It is evident that the prime factors of 42875 can be grouped into triples of equal factors and no factor is left over. Therefore, 42875 is a perfect cube.

This implies that – 42875 is also a perfect cube.

Now, collect one factor from each triplet and multiply, we get: 5 x 7 = 35

This implies that 42875 is a cube of 35. Thus, – 42875 is the cube of – 35.

Q3. Show that the following integers are cubes of negative integers. Also find the integer whose cube is the given integer.

(i) – 5832   (ii) – 2744000

Answer:

In order to check if a negative number is a perfect cube, first check if the corresponding positive integer is a perfect cube. Also, for any positive integer in, – m3 is the cube of – m.

(i) On factorising 5832 into prime factors, we get:

5832 = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3

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

5832 = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3

It is evident that the prime factors of 5832 can be grouped into triples of equal factors and no factor is left over. Therefore, 5832 is a perfect cube.

This implies that -5832 is also a perfect cube.

Now, collect one factor from each triplet and multiply, we get:

2 x 3 x 3 = 18

This implies that 5832 is a cube of 18.

Thus, – 5832 is the cube of – 18.

 

(ii) On factorising 2744000 into prime factors, we get:

2744000 = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5 x 5 x 7 x 7 x 7

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

2744000 = {2 x 2 x 2} x {2 x 2 x 2} x {5 x 5 x 5} x {7 x 7 x 7}

It is evident that the prime factors of 2744000 can be grouped into triples of equal factors and no factor is left over. Therefore, 2744000 is a perfect cube. This implies that – 2744000 is also a perfect cube.

Now, collect one factor from each triplet and multiply, we get: 2 x 2 x 5 x 7 = 140

This implies that 2744000 is a cube of 140. Thus, – 2744000 is the cube of – 140.

 

 

Q4. Find the cube of:

(i) \(\frac{7}{9}\) 

(ii) – \(\frac{8}{11}\)   

(iii) \(\frac{12}{7}\)

(iv) – \(\frac{13}{8}\)  

(v) \(2 \frac{2}{5}\)   

(vi) \(3 \frac{1}{4}\) 

(vii) 0.3 

(viii) 1.5 

(ix) 0.08 

(x) 2.1

Answer:

\((i) \; (\frac{m}{n})^{3} \; = \; \frac{m^{3}}{n^{3}} \\ (\frac{7}{9})^{3} = \frac{7^{3}}{9^{3}} = \frac{7 \times 7 \times 7}{9 \times 9 \times 9} = \frac{343}{729}\)

\((ii) \; (- \frac{m}{n})^{3} \; = \; \frac{- m^{3}}{n^{3}} \\ (- \frac{8}{11})^{3} = – \frac{8^{3}}{11^{3}} = – (\frac{8 \times 8 \times 8}{11 \times 11 \times 11}) = – \frac{512}{1331}\)

\((iii) \; (\frac{m}{n})^{3} \; = \; \frac{m^{3}}{n^{3}} \\ (\frac{12}{7})^{3} = \frac{12^{3}}{7^{3}} = (\frac{12 \times 12 \times 12}{7 \times 7 \times 7}) = \frac{1728}{343}\)

\((iv) \; (- \frac{m}{n})^{3} \; = \; – \frac{m^{3}}{n^{3}} \\ (- \frac{13}{8})^{3} = \frac{13^{3}}{8^{3}} = – (\frac{13 \times 13 \times 13}{8 \times 8 \times 8}) = – \frac{2197}{512}\)

 

(v) We have:

\(2\frac{2}{5} = \frac{12}{5} \\ Also, (\frac{m}{n})^{3} = \frac{m^{3}}{n^{3}} \\ (\frac{12}{5})^{3} = \frac{12^{3}}{5^{3}} = \frac{12 \times 12 \times 12}{5 \times 5 \times 5} = \frac{1728}{125}\)

 

(vi) We have:

\(3\frac{1}{4} = \frac{13}{4} \\ Also, (\frac{m}{n})^{3} = \frac{m^{3}}{n^{3}} \\ (\frac{13}{4})^{3} = \frac{13^{3}}{4^{3}} = \frac{13 \times 13 \times 13}{4 \times 4 \times 4} = \frac{2197}{64}\)

 

(vii) We have:

\(0.3 = \frac{3}{10} \\ Also, (\frac{m}{n})^{3} = \frac{m^{3}}{n^{3}} \\ (\frac{3}{10})^{3} = \frac{3^{3}}{10^{3}} = \frac{3 \times 3 \times 3}{10 \times 10 \times 10} = \frac{27}{1000} = 0.027\)

 

(viii) We have:

\(1.5 = \frac{15}{10} \\ Also, (\frac{m}{n})^{3} = \frac{m^{3}}{n^{3}} \\ (\frac{15}{10})^{3} = \frac{15^{3}}{10^{3}} = \frac{15 \times 15 \times 15}{10 \times 10 \times 10} = \frac{3375}{1000} = 3.375\)

 

(ix) We have:

\(0.08 = \frac{8}{100} \\ Also, (\frac{m}{n}) ^{3} = \frac{m ^{3}}{n ^{3}} \\ (\frac{8}{100}) ^{3} = \frac{8 ^{3}}{100 ^{3}} = \frac{8 \times 8 \times 8}{100 \times 100 \times 100} = \frac{512}{1000000} = 0.000512\)

 

(x) We have:

\(2.1 = \frac{21}{10} \\ Also, (\frac{m}{n}) ^{3} = \frac{m ^{3}}{n ^{3}} \\ (\frac{21}{10}) ^{3} = \frac{21 ^{3}}{10 ^{3}} = \frac{21 \times 21 \times 21}{10 \times 10 \times 10} = \frac{9261}{1000} = 9.261\)

 

 

Q5. Find which of the following numbers are cubes of rational numbers?

(i) \(\frac{27}{64}\)   

(ii) \(\frac{125}{128}\)  

(iii) 0.001331  (iv) 0.04

 Answer:

(i) We have:

\(\frac{27}{64} = \frac{3 \times 3 \times 3}{8 \times 8 \times 8} =\frac{3^{3}}{8^{3}} = (\frac{3}{8})^{3}\)

Therefore, \(\frac{27}{64} \; is \; a \; cube \; of \; \frac{3}{8}\)

 

(ii) We have:

\(\frac{125}{128} = \frac{5 \times 5 \times 5}{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times } =\frac{5^{3}}{2^{3} \times 2^{3} \times 2}\)

It is evident that 128 cannot be grouped into triples of equal factors; Therefore, \(\frac{125}{128}\)   is not a cube of a rational number.

 

(iii) We have:

0.001331 = \(\frac{1331}{1000000} = \frac{11 \times 11 \times 11}{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5} =\frac{11^{3}}{(2 \times 2 \times 5 \times 5)^{3}}\)

 

(iv) We have:

0.04 = \(\frac{4}{100} = \frac{2 \times 2}{5 \times 5 \times 5 \times 5}\)

It is evident that 4 and 100 could not be grouped into triples of equal factors; therefore, 0.04 is not a cube of rational number.


Practise This Question

Find the value of angles 2,3,4.