## RD Sharma Solutions Class 8 Chapter 4 Exercise 4.2

**Q1. Find the cubes of:**

**(i) â€“ 11Â Â Â (ii) – 12Â Â Â Â (iii) – 21**

**Â 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).

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

**(i) â€“ 64Â Â Â **

**(ii) â€“ 1056Â Â **

**(iii) â€“ 2197Â Â **

**(iv) â€“ 2744Â Â Â **

**(v) â€“ 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, – m^{3} is the cube of – m.

**(i)Â Â **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.

**Â **

**(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.

**Â **

**(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.

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**(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.

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**(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) â€“ 5832Â Â (ii) â€“ 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, – m^{3} 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.

**Â **

**(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.

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**Â **

**Q4. Find the cube of:**

**(i) \(\frac{7}{9}\)Â **

**(ii) – \(\frac{8}{11}\)Â Â Â **

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

**(iv) – \(\frac{13}{8}\)Â Â **

**(v) \(2 \frac{2}{5}\) Â Â **

**(vi) \(3 \frac{1}{4}\)Â **

**(vii) 0.3Â **

**(viii) 1.5Â **

**(ix) 0.08Â **

**(x) 2.1**

**Answer:**

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

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

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

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

**Â **

**(v)** We have:

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

**Â **

**(vi)** We have:

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

**Â **

**(vii)** We have:

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

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**(viii)** We have:

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

**Â **

**(ix)** We have:

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

**Â **

**(x)** We have:

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

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**Â **

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

**(i) \(\frac{27}{64}\)Â Â Â **

** (ii) \(\frac{125}{128}\)Â Â **

**(iii) 0.001331Â (iv) 0.04**

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

**Â **

**(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, \(\frac{125}{128}\)

**Â **

**(iii)** We have:

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

**Â **

**(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.