Rationalising a denominator

Question

I have the question "Rationalise the denominator, simplifying where possible. $$\frac {\sqrt{3} }{ \sqrt{2} \left(\sqrt{6} - \sqrt{3} \right)}$$

I am not sure whether I should multiply the top and bottom as it is but reversing the sign so $\sqrt{2}$ ($\sqrt{6}$ + $\sqrt{3}$) or whether I should first expand the brackets of the denominator and then multiply top and bottom by this.

The final answer from looking at the solutions should be $\frac {2 + \sqrt{2}}2$.

Answer 1

$$\frac {\sqrt{3}}{\sqrt{2} \left(\sqrt{6} - \sqrt{3} \right)}\cdot\frac{\sqrt{2} \left(\sqrt{6} + \sqrt{3} \right)}{\sqrt{2} \left(\sqrt{6} +\sqrt{3} \right)}=$$ $$\frac{\sqrt{6}\left(\sqrt{6} +\sqrt{3} \right)}{2(6-3)}=\frac{6+\sqrt{18}}{6}=\frac{6+3\sqrt{2}}{6}=\frac{2+\sqrt{2}}{2}$$

Answer 2

Hint: $\frac{\sqrt{3}}{\sqrt{2}}\frac{1}{\sqrt{6}-\sqrt{3}}=\frac{\sqrt{3}\sqrt{2}}{\sqrt{2}\sqrt{2}}\frac{\sqrt{6}+\sqrt{3}}{\sqrt{6}^2-\sqrt{3}^2}$

Answer 3

Simplifying first may help. Note that $\sqrt 6 = \sqrt 3\cdot\sqrt 2$ $$\frac {\sqrt{3} }{ \sqrt{2} \left(\sqrt{6} - \sqrt{3} \right)} = \frac{\sqrt 3}{\sqrt 2\sqrt 3( \sqrt 2 - 1)} = \frac 1{\sqrt 2 (\sqrt 2 -1)} = \frac 1{2-\sqrt 2}$$

Then simply multiply top and bottom by $2+ \sqrt 2$ to get $$\frac{2 + \sqrt 2}{2^2 - (\sqrt 2)^2} = \frac{2+\sqrt 2}{2}$$