If |x| is so small so that ${\mathrm{x}}^2$ and higher powers of x may be neglected, then an approximate value of $\frac{{\left(1+{\displaystyle \frac{2}{3}}\mathrm{x}\right)}^{-3} {\left(1-15\mathrm{x}\right)}^{-1/5}}{{\left(2-3\mathrm{x}\right)}^4}$ is
 

Given   $={\left(1+\frac{2}{3} \mathrm{x}\right)}^{-3} {\left(1-15\mathrm{x}\right)}^{\frac{-1}{5}} \frac{1}{16} {\left(1-\frac{3}{2} \mathrm{x}\right)}^{-4}$

             $$\begin{gathered} =\left(1-3\left(\frac{2}{3} \mathrm{x}\right)\right) \left(1+15\mathrm{x}.\frac{1}{5}\right) \frac{1}{16} \left(1+4\left(\frac{3}{2}\right)\mathrm{x}\right) \\ =\frac{1}{16} \left(1-2\mathrm{x}\right) (1+3\mathrm{x}) (1+6\mathrm{x}) \\ =\frac{1}{16} (1+\mathrm{x}) (1+6\mathrm{x})=\frac{1}{16} (1+7\mathrm{x}) \end{gathered}$$