Finding the inverse of $2+\sqrt{5}+2\sqrt{7}$ Finding the inverse of $2+\sqrt{5}+2\sqrt{7}$ in the field $\mathbb{Q}(\sqrt{5},\sqrt{7})$.
I know that all the elements of $\mathbb{Q}(\sqrt{5},\sqrt{7})$ are of the form:
$a+b\sqrt{5}+c\sqrt{7}+d\sqrt{35}$, where $a,b,c,d \in \mathbb{Q} $
So I could simply solve: $(2+\sqrt{5}+2\sqrt{7})(a+b\sqrt{5}+c\sqrt{7}+d\sqrt{35})=1$
Which I believe works. Is there a better way of solving this problem? 

\begin{align}
\frac{1}{2+\sqrt{5}+2\sqrt{7}}
&=\frac{2+\sqrt{5}-2\sqrt{7}}{(2+\sqrt{5})^2-28}\\
&=\frac{2+\sqrt{5}-2\sqrt{7}}{-19+4\sqrt{5}}\\
&=\frac{(2+\sqrt{5}-\sqrt{7})(-19-4\sqrt{5})}{361-80}\\
&=\dots
\end{align}
