Answer
$5\sqrt[3]{2}$
Work Step by Step
Factor each radicand so that one factor is a perfect cube:
$\sqrt[3]{27\cdot2}+\sqrt[3]{8\cdot2}$
Recall the property (pg. 367):
$\sqrt[n]{a}*\sqrt[n]{b}=\sqrt[n]{ab}$ (if $\sqrt[n]{a}$ and $\sqrt[n]{b}$ are real numbers)
Applying this property, we get:
$\sqrt[3]{27\cdot2}+\sqrt[3]{8\cdot2}$
$=\sqrt[3]{27}\cdot \sqrt[3]{2}+\sqrt[3]{8}\cdot \sqrt[3]{2}$
Recall that $3^3=27$ and $2^3=8$:
Thus, the expression above simplifies to:
$=3\sqrt[3]{2}+2\sqrt[3]{2}$
$=5\sqrt[3]{2}$
