## Precalculus (6th Edition) Blitzer

$3\sqrt[3]{2}$
RECALL: For any real numbers $a$ and $b$, $\sqrt[3]{a} \cdot \sqrt[3]{b} = \sqrt[3]{ab}.$ Use the rule above to obtain $\sqrt[3]{9(6)} \\=\sqrt{54}.$ Factor the radicand so that one factor is a perfect cube to obtain $\sqrt[3]{27(2)} \\=\sqrt[3]{3^3(2)}.$ Simplify to obtain $3\sqrt[3]{2}.$