## Precalculus (6th Edition)

$\color{blue}{512}$
RECALL: (1) $a^{m/n} = \left(^{1/n}\right)^m$ (2) $a^{1/n} = \sqrt[n]{a}$ (3) For positive real numbers $a$, $\sqrt[n]{a^n}=a$ Use rule (1) above to obtain: $64^{3/2} = \left(64^{1/2}\right)^3$ Use rule (2) above to obtain: $=(\sqrt[2]{64})^3$ With $64=8^2$, the expression above is equivalent to: $=(\sqrt[2]{8^2})^3$ Use rule (3) above to obtain" $=(8)^3 \\=\color{blue}{512}$