Answer
$-2x^{2}\sqrt[5] y$
Work Step by Step
$\sqrt [5] (-32x^{10}y)=\sqrt [5] (-32\times x^{10}\times y)=\sqrt [5] (-32)\times \sqrt[5] (x^{10})\times \sqrt[5] y=-2x^{2}\sqrt[5] y$
We know that $\sqrt[5] (-32)=-2$, because $(-2)^{5}=-32$ and also that $\sqrt[5] (x^{10})=x^{2}$, because $(x^{2})^{5}=x^{2\times5}=x^{10}$.
