Simplify:
$ \left[\left(\frac{1}{3}\right)^{-1}-\left(\frac{2}{5}\right)^{-1}\right]^{-2} \div\left(\frac{3}{4}\right)^{-3} $

Given:

\( \left[\left(\frac{1}{3}\right)^{-1}-\left(\frac{2}{5}\right)^{-1}\right]^{-2} \div\left(\frac{3}{4}\right)^{-3} \)

To do:

We have to simplify \( \left[\left(\frac{1}{3}\right)^{-1}-\left(\frac{2}{5}\right)^{-1}\right]^{-2} \div\left(\frac{3}{4}\right)^{-3} \).

Solution:

We know that,

$a^{-m}=\frac{1}{a^m}$

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

$a^{m}\div a^{n}=a^{m-n}$

Therefore,

$[(\frac{1}{3})^{-1}-(\frac{2}{5})^{-1}]^{-2} \div(\frac{3}{4})^{-3}=[(\frac{3}{1})^{1}-(\frac{5}{2})^{1}]^{-2} \div(\frac{4}{3})^{3}$

$=[3-\frac{5}{2}]^{-2} \times(\frac{3}{4})^{3}$

$=(\frac{6-5}{2})^{-2}\times(\frac{3}{4})^{3}$

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

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

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

$=4\times\frac{27}{64}$

$=\frac{27}{16}$

Hence, $[(\frac{1}{3})^{-1}-(\frac{2}{5})^{-1}]^{-2} \div(\frac{3}{4})^{-3}= \frac{27}{16}$.