How to denest these radicals? I know about the shortcut in which we unnestle radicals but I want to know why it works. For example: How do we unnestle these radicals and solve the problem?
$${1\over \sqrt{7+4\sqrt{3}}} + {1\over \sqrt{7-4\sqrt{3}}}$$
Thanks in advance!
 A: Simple radicals of the form $\sqrt{X\pm Y}$ can be denested with the following identity
$$\sqrt{X\pm Y}=\sqrt{\frac {X+\sqrt{X^2-Y^2}}2}\pm\sqrt{\frac {X-\sqrt{X^2-Y^2}}2}$$
Where $X,Y\in\mathbb{R}$ and a simplification occurs only when $X^2-Y^2$ is a perfect square. Therefore, it follows that$$\sqrt{7\pm4\sqrt3}=\sqrt{\frac {7+\sqrt{49-48}}2}\pm\sqrt{\frac {7-\sqrt{49-48}}2}=2\pm\sqrt3$$Can you complete the rest?


Proof: Assume $\sqrt{X\pm Y}$ to be equal to $\sqrt{A}\pm\sqrt{B}$ and square both sides. Expanding the right-hand side and setting the two sides equal to each other gives our identity.

A: Hint:
$$7\pm4\sqrt3=\pm2\cdot2\cdot\sqrt3+2^2+(\pm\sqrt3)^2=(2\pm\sqrt3)^2$$
Now choose the principal values
A: $(a+b\sqrt{c})^2
=a^2+b^2c+2ab\sqrt{c}
$.
Put values to generate as many examples as you want.
A: $$7-4\sqrt3=(2-\sqrt3)^2$$
and $$7+4\sqrt3=(2+\sqrt3)^2$$
Hence $${1\over \sqrt{7+4\sqrt{3}}} + {1\over \sqrt{7-4\sqrt{3}}}  =\frac{(2+\sqrt3)^2+(2-\sqrt3)^2}{(2+\sqrt3)^2(2-\sqrt3)^2}=\frac{8+6}{(4-3)^2} =\color{red}{14}$$
