## Intermediate Algebra (6th Edition)

$\sqrt [3] 36$
The product rule holds that $\sqrt[n] a\times\sqrt[n] b=\sqrt[n] (ab)$ (where $\sqrt[n] a$ and $\sqrt[n] b$ are real numbers). Therefore, $\sqrt[3] 4\times\sqrt[3] 9=\sqrt[3] (4\times9)=\sqrt [3] 36$