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!

Let $\beta = 2\sqrt{2}-3\sqrt{3}+5$. Then 
$(\beta-5)^2 = 35-12\sqrt{6}$ and $((\beta-5)^2-35)^2 = 864$, and so
$$
\beta^4 - 20 \beta^3 + 80 \beta^2 + 200 \beta = 764
$$
Writing this as
$$
\beta(\beta^3 - 20 \beta^2 + 80 \beta + 200) = 764
$$
gives $\beta^{-1} = \frac{1}{764}(\beta^3 - 20 \beta^2 + 80 \beta + 200)$.
