How to use radical rules for this?

Question

$$\sqrt{3+\sqrt{2\sqrt{7}+1}}-\sqrt{3-\sqrt{2\sqrt{7}+1}}=?$$

Which one of the following is a true answer and why?

1. $\sqrt{7}-7$
2. $1-\sqrt{7}$
3. $\sqrt{7}-1$
4. $\sqrt{7}$
5. $\sqrt{7}+7$

Answer 1

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:

$$\sqrt 7 -1 = \sqrt{\big(\sqrt 7 -1\big)^2} = \sqrt{7 + 1 - 2\sqrt 7} = \sqrt{8 - 2\sqrt 7}$$