Answer
$-2x^{3}y^{4}$
Work Step by Step
$\sqrt[5] (-32x^{15}y^{20})=\sqrt[5] (-32\times x^{15}\times y^{20})=\sqrt[5] (-32)\times\sqrt[5] (x^{15})\times\sqrt[5] (y^{20})=-2x^{3}y^{4}$
$\sqrt[5] (-32)=-2$, because $(-2)^{5}=-32$
$\sqrt[5] (x^{15})=x^{3}$, because $(x^{3})^{5}=x^{3\times5}=x^{15}$
$\sqrt[5] (y^{20})=y^{4}$, because $(y^{4})^{5}=y^{4\times5}=y^{20}$
