A simple question about an expression with square roots

Question

$$\dfrac{\sqrt{6}-\sqrt{3}+3}{3\sqrt{2}+\sqrt{3}+3}$$

I know that we can multiply both the numerator and denominator with $3\sqrt{2}-\sqrt{3}-3$ and then multiply it with $-12-6\sqrt{6}$ this gives the answer. However what I was looking for was more likely a pattern, a common factor between the numerator and denominator, but i failed in finding one.

Answer 1

$$\dfrac{\sqrt{6}-\sqrt{3}+3}{3\sqrt{2}+\sqrt{3}+3}=\frac{\sqrt2-1+\sqrt3}{\sqrt6+1+\sqrt3}=\frac{(\sqrt2-1+\sqrt3)(\sqrt6-1-\sqrt3)}{6-4-2\sqrt3}=$$ $$=\frac{\sqrt3-\sqrt6+\sqrt2-1}{1-\sqrt3}=\frac{\sqrt3(1-\sqrt2)+\sqrt2-1}{1-\sqrt3}=\sqrt2-1.$$ Done!

Answer 2

$$\sqrt6-\sqrt3+3=\sqrt3(\sqrt2-1+\sqrt3)$$

$$3\sqrt2+\sqrt3+3=\sqrt3(\sqrt6+\sqrt3+1)$$

So we can safely cancel out $\sqrt3$

Now multiply the numerator & the denominator by $(\sqrt6+\sqrt3)-1$