Precalculus (10th Edition)

$64, 4, \frac{1}{9}.$
$4^3=4\cdot4\cdot4=64$ Use the rule $a^{m/n}=(\sqrt[n]{a})^m$ to obtain: $8^{2/3}=(\sqrt[3] 8)^2=(\sqrt[3]{2^3})^2=2^24$ Use the rule $a^{-m} = \frac{1}{a^m}, m\gt 0$ to obtain: $3^{-2}=\frac{1}{3^2}=\frac{1}{9}$