Question

Simplify ${\left(\frac{{27}^{-3}}{9^{-3}}\right)}^{\frac{1}{5}}$.

Answer 1

Given: $(\frac{{27}^{-3}}{9^{-3}}{)}^{\frac{1}{5}}$

This can be written as

$=(\frac{27}{9}{)}^{\frac{-3}{5}}$ [Since, $\frac{a^m}{b^m}=(\frac{a}{b}{)}^m$]

$=(3{)}^{\frac{-3}{5}}$

This can be written as

$=(3^3{)}^{\frac{-1}{5}}$

$=(27{)}^{\frac{-1}{5}}$

This can be written as

$=({27}^{-1}{)}^{\frac{1}{5}}$

$=(\frac{1}{27}{)}^{\frac{1}{5}}$

Therefore, $(\frac{{27}^{-3}}{9^{-3}}{)}^{\frac{1}{5}}=(\frac{1}{27}{)}^{\frac{1}{5}}$