## Precalculus (6th Edition) Blitzer

$4$
RECALL: $a^{\frac{m}{n}}=(\sqrt[n]{a})^m.$ Use the rule above with m=2 and n=3 to obtain $(\sqrt[3]{8})^2 \\=(\sqrt[3]{2^3})^2 \\=2^2 \\=4.$