Answer
$-2\sqrt[3]{2}$
Work Step by Step
Factor each radicand so that one factor of the factors is a perfect cube:
$3\sqrt[3]{8\cdot2}-4\sqrt[3]{27\cdot2}+\sqrt[3]{64\cdot2}$
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:
$3\sqrt[3]{8\cdot2}-4\sqrt[3]{27\cdot2}+\sqrt[3]{64\cdot2}$
$=3\sqrt[3]{8}\cdot\sqrt[3]{2}-4\sqrt[3]{27}\cdot \sqrt[3]{2}+\sqrt[3]{64}\cdot\sqrt[3]{2}$
Recall that $2^3=8$, $3^3=27$, and $4^3=64$.
Thus, the expression above simplifies to:
$=3\cdot2\sqrt[3]{2}-4\cdot3\sqrt[3]{2}+4\cdot\sqrt[3]{2}$
$=6\sqrt[3]{2}-12\sqrt[3]{2}+4*\sqrt[3]{2}$
$=(6-12+4)\sqrt[3]{2}$
$=-2\sqrt[3]{2}$
