Rationalizing the fraction $\frac{1}{1-\sqrt2 -\sqrt3}$

Question

I'm having problem in rationalizing the following root with the fraction $$\frac{1}{1-\sqrt2 -\sqrt3}$$ Eventually after many tries, I found the solution which was : $$\frac{-\sqrt2 (1-\sqrt2 +\sqrt3 )}{4}$$ But I want to know if there's a specific method to use in a case like this. Thanks

Answer 1

$$\frac{1}{1-(\sqrt2 +\sqrt3)}=\frac{1}{1-(\sqrt2 +\sqrt3)}\frac{1+(\sqrt2 +\sqrt3)}{1+(\sqrt2 +\sqrt3)}=$$ $$=\frac{1+(\sqrt2 +\sqrt3)}{1-(5 +2\sqrt6)}=\frac{1+(\sqrt2 +\sqrt3)}{2(\sqrt6-2)}=$$ $$=\frac{1+(\sqrt2 +\sqrt3)}{2(\sqrt6-2)}\frac{\sqrt6+2}{\sqrt6+2}=\frac{(1+\sqrt2 +\sqrt3)(\sqrt2 +\sqrt3)}{64}$$

Answer 2

When I rationalize I always admire the property $$(a-b)(a+b)=a^2-b^2$$ Using this fact try stating $a=1$ and $b=\sqrt{2}-\sqrt{3}$. You will get another situation where you are going to use this property again but to answer your question if you have square root in your denominator then try to put it in the form of $(a-b)$ form and finally multiply both the numerator and denominator by $(a + b)$.