Solve $\displaystyle \frac{3}{\sqrt{2}} \times (2\sqrt{3}-6\sqrt{3})$
First, I rationalize the first fraction and get $\displaystyle \frac{3\sqrt{2}}{2}$ and then multiply to get $\displaystyle \frac{-9\sqrt{6}}{2}$
Is this correct?
$\displaystyle \begin{aligned}
\dfrac{3}{\sqrt{2}} \times (2\sqrt{3}-6\sqrt{3}) &= \dfrac{3}{\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}}(-4\sqrt{3}) \\
&= \dfrac{3\sqrt{2}}{2} \cdot (-4\sqrt{3}) \\
&= \dfrac{12\sqrt{6}}{2} \\
&= -6\sqrt{6}
\end{aligned}$
EDIT: Miscopied the problem...