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.

To get rid of the square roots of the denominator, you may use $a^2-b^2=(a+b)(a-b)$.
$$\dfrac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}\\=\dfrac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{(\sqrt{7}-2\sqrt{5})^2-3}\\=\dfrac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{7-4\sqrt{35}+20-3}\\=\dfrac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{24-4\sqrt{35}}\\=\dfrac{(\sqrt{7}-2\sqrt{5}-\sqrt{3})(24+4\sqrt{35})}{24^2-16*35}\\=\dfrac{(\sqrt{7}-2\sqrt{5}-\sqrt{3})(24+4\sqrt{35})}{16}\\=\dfrac{-24\sqrt3-20\sqrt5-16\sqrt7-4\sqrt{105}}{16}\\=\dfrac{-6\sqrt3-5\sqrt5-4\sqrt7-\sqrt{105}}{4}$$
