## Intermediate Algebra (6th Edition)

$\frac{1}{16}$
$64^{-\frac{2}{3}}=\frac{1}{64^{\frac{2}{3}}}$ We know that $a^{\frac{1}{n}}=\sqrt[n] a$ (where n is positive integer greater than 1 and $\sqrt[n] a$ is a real number). Therefore, $\frac{1}{64^{\frac{2}{3}}}=\frac{1}{\sqrt[3] (64^{2})}=\frac{1}{\sqrt[3] ((8^{2})^{2})}=\frac{1}{\sqrt[3] (8\times8\times8\times8)}=\frac{1}{\sqrt[3] 8\times\sqrt[3] 8\times\sqrt[3] 8\times\sqrt[3] 8}=\frac{1}{2\times2\times2\times2}=\frac{1}{16}$