Question

Fill in the blanks.

${\left(\frac{81}{16}\right)}^{-\frac{3}{4}}\times {\left(\frac{25}{9}\right)}^{-\frac{3}{2}}=$

Answer 1

Simplify the given expression using the law of radical exponent.

$\begin{array}{l}{\left(\frac{81}{16}\right)}^{-\frac{3}{4}}\times {\left(\frac{25}{9}\right)}^{-\frac{3}{2}}\\ ={\left(\frac{3^4}{2^4}\right)}^{-\frac{3}{4}}\times {\left(\frac{5^2}{3^2}\right)}^{-\frac{3}{2}}\left[\because \frac{a^m}{b^m}={\left(\frac{a}{b}\right)}^m\right]\\ ={\left\{{\left(\frac{3}{2}\right)}^4\right\}}^{-\frac{3}{4}}\times {\left\{{\left(\frac{5}{3}\right)}^2\right\}}^{-\frac{3}{2}}\left[\because {\left(a^m\right)}^n=a^{mn}\right]\\ ={\left(\frac{3}{2}\right)}^{4\times -\frac{3}{4}}\times {\left(\frac{5}{3}\right)}^{2\times -\frac{3}{2}}\\ ={\left(\frac{3}{2}\right)}^{-3}\times {\left(\frac{5}{3}\right)}^{-3}\\ =\frac{{\left(3\right)}^{-3}}{{\left(2\right)}^{-3}}\times \frac{{\left(5\right)}^{-3}}{{\left(3\right)}^{-3}}\left[\because \frac{a^m}{b^m}={\left(\frac{a}{b}\right)}^m\right]\\ =\frac{{\left(5\right)}^{-3}}{{\left(2\right)}^{-3}}\\ =\frac{2^3}{5^3}\left[\because a^{-n}=\frac{1}{a^n}\right]\\ =\frac{8}{125}\end{array}$

Hence, ${\left(\frac{81}{16}\right)}^{-\frac{3}{4}}\times {\left(\frac{25}{9}\right)}^{-\frac{3}{2}}=\underset{¯}{\frac{8}{125}}$