# Taking Radicals Out of Denominator

How does one take radicals out of the denominator of

$$\frac{1}{a + b(\sqrt{2} + \sqrt{3})}$$

assuming $a,b \in \mathbb{Q}$ not both equal to $0$. I know the trick is to multiply this expression by $1$ s.t. the radicals are removed, but I can't think of such $c/d = 1$ to try.

Try

$$\frac{ a + b \sqrt{2} - b \sqrt{3} } { a + b \sqrt{2} - b \sqrt{3} } \times \frac{ a - b \sqrt{2} + b \sqrt{3} } { a - b \sqrt{2} + b \sqrt{3} } \times \frac{ a - b \sqrt{2} - b \sqrt{3} } { a - b \sqrt{2} - b \sqrt{3} } .$$

The idea is that you have to multiply by all possible conjugates.

Hint: Multiply numerator and denominator first by $a-b(\sqrt2+\sqrt3)$, to get in the denominator a term of form $c + d\sqrt6$. Now you know the next step.

This is just like the usual method for rationalizing the denominator, only you need $4$ factors, not two, to get the rational number you are looking for. That is, multiply the top and the bottom by $a+b(\sqrt2-\sqrt3)$, then by $a+b(-\sqrt2+\sqrt3)$ then by $a+b(-\sqrt2-\sqrt3)$.

You can check directly that $$[a+b(\sqrt2+\sqrt3)][a+b(\sqrt2-\sqrt3)][a+b(-\sqrt2+\sqrt3)][a+b(-\sqrt2-\sqrt3)]$$ is a rational number.

Readers who have seen some Galois theory will recognize this as the norm in a Klein-4 extension.