Answer
$33\sqrt{2}$
Work Step by Step
Factor each radicand so that one factor is a perfect square:
$6\sqrt{9\cdot2}+3\sqrt{25\cdot2}$
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:
$6\sqrt{9\cdot2}+3\sqrt{25\cdot2}$
$=6\sqrt{9} \cdot \sqrt{2}+3\sqrt{25} \cdot \sqrt{2}$
$=6\cdot 3\sqrt{2}+3\cdot5\sqrt{2}$
$=18\sqrt{2}+15\sqrt{2}$
$=33\sqrt{2}$
