Simplify:
(i) $ \left\{\left(\frac{1}{3}\right)^{-3}-\left(\frac{1}{2}\right)^{-3}\right\} \div\left(\frac{1}{4}\right)^{-3} $
(ii) $ \left(3^{2}-2^{2}\right) \times\left(\frac{2}{3}\right)^{-3} $
(iii) $ \left\{\left(\frac{1}{2}\right)^{-1} \times(-4)^{-1}\right\}^{-1} $
(iv) $ \left[\left\{\left(\frac{-1}{4}\right)^{2}\right\}^{-2}\right]^{-1} $
(v) $ \left\{\left(\frac{2}{3}\right)^{2}\right\}^{3} \times\left(\frac{1}{3}\right)^{-4} \times 3^{-1} \times 6^{-1} $

To do:

We have to simplify given rational numbers.

Solution:

We know that,

$\frac{1}{a^m}=a^{-m}$ 

$a^{m}=(\frac{1}{a})^{-m}$ 

$(\frac{a}{b})^m=(\frac{b}{a})^{-m}$

$a^m \times a^n=a^{m+n}$

$(a^m)^n=a^{mn}$

Therefore,

(i) $[(\frac{1}{3})^{-3}-(\frac{1}{2})^{-3}] \div(\frac{1}{4})^{-3}=(3^{3}-2^{3}) \div(4)^{3}$

$=(27-8) \div 64$

$=\frac{19}{64}$

(ii) $(3^{2}-2^{2}) \times(\frac{2}{3})^{-3}=(3^{2}-2^{2}) \times(\frac{3}{2})^{3}$

$=(9-4) \times \frac{27}{8}$

$=\frac{5\times27}{8}$

$=\frac{135}{8}$

(iii) $((\frac{1}{2})^{-1} \times(-4)^{-1})^{-1}=((2)^{1} \times(\frac{1}{-4})^{1})^{-1}$

$=(2 \times \frac{1}{-4})^{-1}$

$=(\frac{1}{-2})^{-1}$

$=(-2)^1$

$=-2$

(iv) $[\{(\frac{-1}{4})^{2}\}^{-2}]^{-1}=[(\frac{-1}{4})^{2 \times(-2)}]^{-1}$

$=(\frac{-1}{4})^{-4\times(-1)}$

$=(\frac{-1}{4})^4$

$=\frac{(-1)^4}{4^4}$

$=\frac{1}{256}$ 

(v) $\{(\frac{2}{3})^{2}\}^{3} \times(\frac{1}{3})^{-4} \times 3^{-1} \times 6^{-1}=(\frac{2}{3})^{2 \times 3} \times(\frac{3}{1})^{4} \times \frac{1}{3^{1}} \times \frac{1}{6^{1}}$

$=(\frac{2}{3})^{6} \times(3)^{4} \times \frac{1}{3} \times \frac{1}{6}$

$=\frac{2 \times 2 \times 2 \times 2 \times 2 \times 2}{3 \times 3 \times 3 \times 3 \times 3 \times 3} \times 3 \times 3 \times 3 \times 3 \times \frac{1}{3}\times \frac{1}{6}$

$=\frac{64}{3 \times 3 \times 3 \times 6}$

$=\frac{64}{81 \times 2}$

$=\frac{32}{81}$

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Updated on: 10-Oct-2022

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