RD Sharma solution class 8 chapter 4 Cube and Cube roots Exercise 4.2

Exercise 4.2

Page-4.13

Question 1:

Find the cubes of:
(i) −11
(ii) −12
(iii) −21

Answer 1:

(i)
Cube of $-$11 is given as:
${\left(-11\right)}^3=-11\times -11\times -11=-1331$
Thus, the cube of 11 is ($-$1331).

(ii)
Cube of $-$12 is given as:  
${\left(-12\right)}^3=-12\times -12\times -12=-1728$

Thus, the cube of $-$12 is ($-$1728).

(iii)
Cube of $-$21 is given as:  
${\left(-21\right)}^3=-21\times -21\times -21=-9261$

Thus, the cube of $-$21 is ($-$9261).

Question 2:

Which of the following numbers are cubes of negative integers
(i) −64
(ii) −1056
(iii) −2197
(iv) −2744
(v) −42875

Answer 2:

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 factorising 64 into prime factors, we get:
$64=2\times 2\times 2\times 2\times 2\times 2$
On grouping the factors in triples of equal factors, we get:
$64=\left\{2\times 2\times 2\right\}\times \left\{2\times 2\times 2\right\}$
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: 
$2\times 2=4$ 
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:
$1056=2\times 2\times 2\times 2\times 2\times 3\times 11$
On grouping the factors in triples of equal factors, we get:​
$1056=\left\{2\times 2\times 2\right\}\times 2\times 2\times 3\times 11$
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:
$2197=13\times 13\times 13$
On grouping the factors in triples of equal factors, we get:​
$2197=\left\{13\times 13\times 13\right\}$
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 factorising 2744 into prime factors, we get:
$2744=2\times 2\times 2\times 7\times 7\times 7$
On grouping the factors in triples of equal factors, we get:​
$2744=\left\{2\times 2\times 2\right\}\times \left\{7\times 7\times 7\right\}$
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: 
$2\times 7=14$
This implies that 2744 is a cube of 14.
Thus, $-$2744 is the cube of $-$14.

(v)
On factorising 42875 into prime factors, we get:
$42875=5\times 5\times 5\times 7\times 7\times 7$
On grouping the factors in triples of equal factors, we get:​
$42875=\left\{5\times 5\times 5\right\}\times \left\{7\times 7\times 7\right\}$
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: 
$5\times 7=35$
This implies that 42875 is a cube of 35.
Thus, $-$42875 is the cube of $-$35.

Question 3:

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 3:

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 factorising 5832 into prime factors, we get:
$5832=2\times 2\times 2\times 3\times 3\times 3\times 3\times 3\times 3$
On grouping the factors in triples of equal factors, we get:
$5832=\left\{2\times 2\times 2\right\}\times \left\{3\times 3\times 3\right\}\times \left\{3\times 3\times 3\right\}$
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:
$2\times 3\times 3=18$ 
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:
$2744000=2\times 2\times 2\times 2\times 2\times 2\times 5\times 5\times 5\times 7\times 7\times 7$
On grouping the factors in triples of equal factors, we get:​
$2744000=\left\{2\times 2\times 2\right\}\times \left\{2\times 2\times 2\right\}\times \left\{5\times 5\times 5\right\}\times \left\{7\times 7\times 7\right\}$
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: 
$2\times 2\times 5\times 7=140$ 
This implies that 2744000 is a cube of 140.
Thus, $-$2744000 is the cube of $-$140.

Question 4:

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 4:

(i)
$\because$ ${\left(\frac{m}{n}\right)}^3=\frac{m^3}{n^3}$

$\therefore$ ${\left(\frac{7}{9}\right)}^3=\frac{7^3}{9^3}=\frac{7\times 7\times 7}{9\times 9\times 9}=\frac{343}{729}$

(ii)
$\because$ ${\left(-\frac{m}{n}\right)}^3=-\frac{m^3}{n^3}$ 

$\therefore$ ${\left(-\frac{8}{11}\right)}^3=-{\left(\frac{8}{11}\right)}^3=-\left(\frac{8^3}{{11}^3}\right)=-\left(\frac{8\times 8\times 8}{11\times 11\times 11}\right)=-\frac{512}{1331}$

(iii)
$\because$ ${\left(\frac{m}{n}\right)}^3=\frac{m^3}{n^3}$

$\therefore$ ${\left(\frac{12}{7}\right)}^3=\frac{{12}^3}{7^3}=\frac{12\times 12\times 12}{7\times 7\times 7}=\frac{1728}{343}$

(iv)
$\because$ ${\left(-\frac{m}{n}\right)}^3= -\frac{m^3}{n^3}$

$\therefore$ ${\left(-\frac{13}{8}\right)}^3=-{\left(\frac{13}{8}\right)}^3=-\left(\frac{{13}^3}{8^3}\right)=-\left(\frac{13\times 13\times 13}{8\times 8\times 8}\right)=-\frac{2197}{512}$

(v)
We have:

$2\frac{2}{5}=\frac{12}{5}$

Also, ${\left(\frac{m}{n}\right)}^3=\frac{m^3}{n^3}$

$\therefore$ ${\left(\frac{12}{5}\right)}^3=\frac{{12}^3}{5^3}=\frac{12\times 12\times 12}{5\times 5\times 5}=\frac{1728}{125}$

(vi)
We have:

$3\frac{1}{4}=\frac{13}{4}$

Also, ${\left(\frac{m}{n}\right)}^3=\frac{m^3}{n^3}$
$\therefore$ ${\left(\frac{13}{4}\right)}^3=\frac{{13}^3}{4^3}=\frac{13\times 13\times 13}{4\times 4\times 4}=\frac{2197}{64}$

(vii)
We have:

$0.3=\frac{3}{10}$

Also, ${\left(\frac{m}{n}\right)}^3=\frac{m^3}{n^3}$

$\therefore$ ${\left(\frac{3}{10}\right)}^3=\frac{3^3}{{10}^3}=\frac{3\times 3\times 3}{10\times 10\times 10}=\frac{27}{1000}=0.027$

(viii)
We have:

$1.5=\frac{15}{10}$

Also, ${\left(\frac{m}{n}\right)}^3=\frac{m^3}{n^3}$

$\therefore$ ${\left(\frac{15}{10}\right)}^3=\frac{{15}^3}{{10}^3}=\frac{15\times 15\times 15}{10\times 10\times 10}=\frac{3375}{1000}=3.375$

(ix)
We have:

$0.08=\frac{8}{100}$

Also, ${\left(\frac{m}{n}\right)}^3=\frac{m^3}{n^3}$

$\therefore$ ${\left(\frac{8}{100}\right)}^3=\frac{8^3}{{100}^3}=\frac{8\times 8\times 8}{100\times 100\times 100}=\frac{512}{1000000}=0.000512$

(x)
We have:

$2.1=\frac{21}{10}$

Also, ${\left(\frac{m}{n}\right)}^3=\frac{m^3}{n^3}$

$\therefore$ ${\left(\frac{21}{10}\right)}^3=\frac{{21}^3}{{10}^3}=\frac{21\times 21\times 21}{10\times 10\times 10}=\frac{9261}{1000}=9.261$

Question 5:

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 5:

(i)
We have:

$\frac{27}{64}=\frac{3\times 3\times 3}{8\times 8\times 8}=\frac{3^3}{8^3}={\left({\displaystyle \frac{3}{8}}\right)}^3$

Therefore, $\frac{27}{64}$ is a cube of $\frac{3}{8}$.

(ii)
We have:

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

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:

$0.001331=\frac{1331}{1000000}=\frac{11\times 11\times 11}{2\times 2\times 2\times 2\times 2\times 2\times 5\times 5\times 5\times 5\times 5\times 5}=\frac{{11}^3}{{\left(2\times 2\times 5\times 5\right)}^3}=\frac{{11}^3}{{100}^3}={\left({\displaystyle \frac{11}{100}}\right)}^3$

Therefore, $0.001331$ is a cube of $\frac{11}{100}$.

(iv)
We have:

$0.04=\frac{4}{100}=\frac{2\times 2}{2\times 2\times 5\times 5}$

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