Question

Solve this:
Pontential form:${\left(\frac{1}{81}\right)}^{\frac{1}{4}}-{\left(\frac{27}{64}\right)}^{\frac{-1}{3}}÷{\left(\frac{1}{8}\right)}^{\frac{-2}{3}}+{\left(\frac{1}{16}\right)}^{\frac{-3}{4}}$

Answer 1

Dear Student,

$$\begin{gathered} {\left(\frac{1}{81}\right)}^{\frac{1}{4}}-{\left(\frac{27}{64}\right)}^{\frac{-1}{3}}÷{\left(\frac{1}{8}\right)}^{\frac{-2}{3}}+{\left(\frac{1}{16}\right)}^{\frac{-3}{4}} \\ ={\left({\left(\frac{1}{3}\right)}^4\right)}^{\frac{1}{4}}-{\left({\left(\frac{3}{4}\right)}^3\right)}^{\frac{-1}{3}}÷{\left({\left(\frac{1}{2}\right)}^3\right)}^{\frac{-2}{3}}+{\left({\left(\frac{1}{2}\right)}^4\right)}^{\frac{-3}{4}} \\ ={\left(\frac{1}{3}\right)}^{4\times \frac{1}{4}}-{\left(\frac{3}{4}\right)}^{3\times \frac{-1}{3}}÷{\left(\frac{1}{2}\right)}^{3\times \frac{-2}{3}}+{\left(\frac{1}{2}\right)}^{4\times \frac{-3}{4}}\left[\because {\left(a^m\right)}^n=a^{mn}\right] \\ =\frac{1}{3}-{\left(\frac{3}{4}\right)}^{-1}÷{\left(\frac{1}{2}\right)}^{-2}+{\left(\frac{1}{2}\right)}^{-3} \\ =\frac{1}{3}-\left(\frac{4}{3}\times 2^{-2}\right)+2^3\left[\because a^{-m}=\frac{1}{a^m}\right] \\ =\frac{1}{3}-\left(\frac{4}{3}\times \frac{1}{4}\right)+2^3\left[\because a^{-m}=\frac{1}{a^m}\right] \\ =\frac{1}{3}-\frac{1}{3}+8=8 \end{gathered}$$​