Inverse element in field extension $\mathbb{Q}(\sqrt{2},\sqrt{3})$ Determine $(2\sqrt{2}-3\sqrt{3}+5)^{-1}$ in $\mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$, which is of degree $4$ over $\mathbb{Q}$ and for $a= \sqrt{2} + \sqrt{3}$ follows $a^4-10a^2+1=0$.
I did calculate inverse elements in field extensions of degree $2$ over $\mathbb{Q}$, but am somehow stuck with this problem, any help is very much appreciated!

You can carry out the "rationalise the denominator" in two steps.
First multiply $\frac{1}{5+2\sqrt{2}-3\sqrt{3}}$ by $\frac{5+2\sqrt{2}+3\sqrt{3}}{5+2\sqrt{2}+3\sqrt{3}}$ and get $\frac{5+2\sqrt{2}+3\sqrt{3}}{6+20\sqrt{2}}$.
Now multiply by by $\frac{6-20\sqrt{2}}{6-20\sqrt{2}}$ and get
$\frac{(5+2\sqrt{2}+3\sqrt{3})(6-20\sqrt{2})}{-764}$.
Multiply out the numerator and you're done.
