Answer
$\sqrt[5]{\dfrac{32}{m^{6}n^{13}}}=\dfrac{2\sqrt[5]{m^{4}n^{2}}}{m^{2}n^{3}}$
Work Step by Step
$\sqrt[5]{\dfrac{32}{m^{6}n^{13}}}$
Rewrite this expression as $\dfrac{\sqrt[5]{32}}{\sqrt[5]{m^{6}n^{13}}}$ and simplify it:
$\sqrt[5]{\dfrac{32}{m^{6}n^{13}}}=\dfrac{\sqrt[5]{32}}{\sqrt[5]{m^{6}n^{13}}}=\dfrac{2}{mn^{2}\sqrt[5]{mn^{3}}}=...$
Multiply the fraction by $\dfrac{\sqrt[5]{m^{4}n^{2}}}{\sqrt[5]{m^{4}n^{2}}}$ and simplify:
$...=\dfrac{2}{mn^{2}\sqrt[5]{mn^{3}}}\cdot\dfrac{\sqrt[5]{m^{4}n^{2}}}{\sqrt[5]{m^{4}n^{2}}}=\dfrac{2\sqrt[5]{m^{4}n^{2}}}{mn^{2}\sqrt[5]{m^{5}n^{5}}}=\dfrac{2\sqrt[5]{m^{4}n^{2}}}{mn^{2}(m)(n)}=...$
$...=\dfrac{2\sqrt[5]{m^{4}n^{2}}}{m^{2}n^{3}}$
