How to use radical rules for this? 
$$\sqrt{3+\sqrt{2\sqrt{7}+1}}-\sqrt{3-\sqrt{2\sqrt{7}+1}}=?$$
Which one of the following is a true answer and why?

*

*$\sqrt{7}-7$

*$1-\sqrt{7}$

*$\sqrt{7}-1$

*$\sqrt{7}$

*$\sqrt{7}+7$


Square it and take root:
\begin{align}
& \sqrt{3+\sqrt{2\sqrt{7}+1}}-\sqrt{3-\sqrt{2\sqrt{7}+1}} \\
= & \sqrt{\Big[\sqrt{3+\sqrt{2\sqrt{7}+1}}-\sqrt{3-\sqrt{2\sqrt{7}+1}}\Big]^2} \\
= & \sqrt{\big(3+\sqrt{2\sqrt{7}+1}\big) + \big(3-\sqrt{2\sqrt{7}+1}\big) - 2\sqrt{(3+\sqrt{2\sqrt{7}+1})(3-\sqrt{2\sqrt{7}+1})}} \\
= & \sqrt{6 - 2\sqrt{9-(2\sqrt{7}+1)}} \\
= & \sqrt{6 - 2\sqrt{8-2\sqrt 7}} \\
= & \sqrt{6 - 2(\sqrt 7 -1)} \\
= & \sqrt{8 - 2\sqrt 7} \\
= & \sqrt 7 -1
\end{align}
where I have used, twice, the fact that $\sqrt{8-2\sqrt 7} = \sqrt 7 -1$. This can be verified by, again, squaring then taking square root:
$$\sqrt 7 -1 = \sqrt{\big(\sqrt 7 -1\big)^2} = \sqrt{7 + 1 - 2\sqrt 7} = \sqrt{8 - 2\sqrt 7}$$
