Question

Simplify: ${\left(\frac{81}{16}\right)}^{\frac{-3}{4}}\times \left\{{\left(\frac{25}{9}\right)}^{\frac{-3}{2}}÷{\left(\frac{5}{2}\right)}^{-3}\right\}$

Answer 1

${\left(\frac{81}{16}\right)}^{\frac{-3}{4}}\times \left({\left(\frac{25}{9}\right)}^{\frac{-3}{2}}÷{\left(\frac{5}{2}\right)}^{-3}\right)$

$\Rightarrow$${\left(\frac{9\times 9}{4\times 4}\right)}^{\frac{-3}{4}}\times \left({\left(\frac{5\times 5}{3\times 3}\right)}^{\frac{-3}{2}}÷{\left(\frac{2}{5}\right)}^3\right)$

$\Rightarrow$${\left(\frac{3}{2}\right)}^{4\times \frac{-3}{4}}\times \left({\left(\frac{5}{3}\right)}^{\frac{-3}{2}\times 2}\times {\left(\frac{5}{2}\right)}^3\right)$

$={\left(\frac{3}{2}\right)}^{-3}\times \left({\left(\frac{3}{5}\right)}^3\times {\left(\frac{5}{2}\right)}^3\right)$

$={\left(\frac{3}{2}\right)}^{-3}\times {\left(\frac{3}{2}\right)}^3$

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