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) \( 2.1 \)

To find: 

We have to find the cube of the given numbers.

Solution:

(i) $(\frac{7}{9})^{3}=\frac{7^3}{9^3}$

$=\frac{7 \times 7 \times 7}{9 \times 9 \times 9}$

$=\frac{343}{729}$

(ii) $(\frac{-8}{11})^{3}=\frac{(-8)^3}{(11)^3}$

$=\frac{-8 \times -8 \times -8}{11 \times 11 \times 11}$

$=\frac{-512}{1331}$

(iii) $(\frac{12}{7})^{3}=\frac{(12)^3}{(7)^3}$

$=\frac{12 \times 12 \times 12}{7 \times 7 \times 7}$

$=\frac{1728}{343}$

(iv) $(\frac{-13}{8})^{3}=\frac{(-13)^3}{(8)^3}$

$=\frac{-13 \times -13 \times -13}{8 \times 8 \times 8}$

$=\frac{-2197}{512}$

(v) $2 \frac{2}{5}=\frac{2\times5+2}{5}$

$=\frac{12}{5}$

$(\frac{12}{5})^{3}=\frac{(12)^3}{(5)^3}$

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

$=\frac{1728}{125}$

(vi) $3 \frac{1}{4}=\frac{3\times4+1}{4}$

$=\frac{13}{4}$

$(\frac{13}{4})^{3}=\frac{(13)^3}{(4)^3}$

$=\frac{13 \times 13 \times 13}{4 \times 4 \times 4}$

$=\frac{2197}{64}$ 

(vii) $0.3=\frac{3}{10}$

$(\frac{3}{10})^{3}=\frac{(3)^3}{(10)^3}$

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

$=\frac{27}{1000}$ 

$=0.027$

(viii) $1.5=\frac{15}{10}$

$(\frac{15}{10})^{3}=\frac{(15)^3}{(10)^3}$

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

$=\frac{3375}{1000}$ 

$=3.375$

(ix) $0.08=\frac{8}{100}$

$(\frac{8}{100})^{3}=\frac{(8)^3}{(100)^3}$

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

$=\frac{512}{1000000}$ 

$=0.000512$

(x) $2.1=\frac{21}{10}$

$(\frac{21}{10})^{3}=\frac{(21)^3}{(10)^3}$

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

$=\frac{9261}{1000}$ 

$=9.261$

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Updated on: 10-Oct-2022

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