Answer
$-2b^{2}\sqrt[5] 2$
Work Step by Step
$\sqrt[5] (-64b^{10})=\sqrt[5] (-32\times b^{10}\times2)=\sqrt[5] (-32)\times \sqrt[5] (b^{10})\times\sqrt[5] 2=-2b^{2}\sqrt[5] 2$
We know that $\sqrt[5] (-32)=-2$, because $(-2)^{5}=-32$.
We know that $\sqrt[5] (b^{10})=b^{2}$, because $(b^{2})^{5}=b^{2\times5}=b^{10}$.
