Answer
$9\sqrt[3]{3}-6\sqrt[3]{2}$
Work Step by Step
Factor each radicand so that one factor is a perfect cube:
$3\sqrt[3]{27\cdot 3}-2\sqrt[3]{27\cdot2}$
Recall the property (pg. 367):
$\sqrt[n]{a}\cdot\sqrt[n]{b}=\sqrt[n]{ab}$ (if $\sqrt[n]{a}$ and $\sqrt[n]{b}$ are real numbers)
Applying this property, we get:
$3\sqrt[3]{27}\cdot \sqrt[3]{3}-2\sqrt[3]{27}\cdot \sqrt[3]{2}$
Recall that $3^3=27$:
Thus, the expression above simplifies to:
$3\cdot 3\sqrt[3]{3}-2\cdot 3\sqrt[3]{2}$
$=9\sqrt[3]{3}-6\sqrt[3]{2}$
