Get rid of the square roots of the denominator: $\dfrac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}$ > How to get rid of the square roots of the denominator: $\dfrac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}$? * I squared the whole denominator, but that didn't help. * Also I searched for a propriety or identity like $A^2-B^2$, but I didn't see one that could fit. Any help is appreciated.

$$\begin{align}\frac{1}{\sqrt 7-2\sqrt 5+\sqrt 3}&=\frac{1}{\sqrt 7-2\sqrt 5+\sqrt 3}\cdot \frac{\sqrt 7-2\sqrt 5-\sqrt 3}{\sqrt 7-2\sqrt 5-\sqrt 3}\\\&=\frac{\sqrt 7-2\sqrt 5-\sqrt 3}{(\sqrt 7-2\sqrt 5)^2-3}\\\&=\frac{\sqrt 7-2\sqrt 5-\sqrt 3}{4(6-\sqrt{35})}\cdot\frac{6+\sqrt{35}}{6+\sqrt{35}}\\\&=\frac{(\sqrt 7-2\sqrt 5-\sqrt 3)(6+\sqrt{35})}{4}\end{align}$$