Question

Solve this:
Q. Simplify $\frac{{\displaystyle {\left(25\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\displaystyle \times }{\displaystyle {{\displaystyle \left(243\right)}}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}}}{{\left(125\right)}^{{\displaystyle \raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\times {\left(81\right)}^{{\displaystyle \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}}$

Answer 1

Dear Student,

$$\begin{gathered} \mathit{Q}.\frac{{\left(25\right)}^{{\displaystyle \frac{-3}{2}}}\times {\left(243\right)}^{{\displaystyle \frac{-3}{5}}}}{{\left(125\right)}^{{\displaystyle \frac{-5}{3}}}\times {\left(81\right)}^{{\displaystyle \frac{-3}{4}}}}=\frac{5^{2\times {\displaystyle \frac{-3}{2}}}\times 3^{5\times {\displaystyle \frac{-3}{5}}}}{5^{3\times {\displaystyle \frac{-5}{3}}}\times 3^{4\times {\displaystyle \frac{-3}{4}}}} \\ =\frac{5^{-3}\times 3^{-3}}{5^{-5}\times 3^{-3}}\left[\frac{1}{a^{-b}}=a^b\right] \\ =5^{-3}\times 3^{-3}\times 5^5\times 3^3\left[a^b\times a^c=a^{b+c}\right] \\ =5^2\times 1 \\ =25Ans \end{gathered}$$

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