Question

${\left(125\right)}^{\frac{-2}{3}}÷{\left(32\right)}^{\frac{-2}{5}}$

Answer 1

Now,

${\left(125\right)}^{\frac{-2}{3}}÷{\left(32\right)}^{\frac{-2}{5}}$

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

$=\frac{5^{-2}}{2^{-2}}$

$=\frac{4}{25}$

$=\frac{16}{100}$

$=0.16$.