Answer
$-2x^{2}y$
Work Step by Step
$\sqrt[5] (-32x^{10}y^{5})=-2x^{2}y$, because
$(-2x^{2}y)^{5}=-2x^{2}y\times -2x^{2}y\times -2x^{2}y\times -2x^{2}y\times -2x^{2}y=(-2\times-2\times-2\times-2\times-2)\times x^{2+2+2+2+2}\times y^{1+1+1+1+1}=-32x^{10}y^{5}$
