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? 

Hint: Consider the product $(2+\sqrt{5}+2\sqrt{7})(2-\sqrt{5}+2\sqrt{7})(2+\sqrt{5}-2\sqrt{7})(2-\sqrt{5}-2\sqrt{7})$.
