## Intermediate Algebra (6th Edition)

$-\frac{4a}{b^{3}}$
$\sqrt[3] (\frac{64a^{3}}{b^{9}})=\frac{4a}{b^{3}}$, because $(\frac{4a}{b^{3}})^{3}=\frac{4a}{b^{3}}\times\frac{4a}{b^{3}}\times \frac{4a}{b^{3}}=\frac{(4\times4\times4)\times a^{1+1+1}}{b^{3+3+3}}=\frac{64a^{3}}{b^{9}}$ Therefore, $-\sqrt[3] (\frac{64a^{3}}{b^{9}})=-\frac{4a}{b^{3}}$. We are just taking the opposite of $\sqrt[3] (\frac{64a^{3}}{b^{9}})$.