The square of the solution of the equation... The square of the solution of the equation 
$x\sqrt{7} + \sqrt{8 - 3\sqrt{7}} - \sqrt{8 + 3\sqrt{7}} = 0$
is equal to: ...
$x\sqrt{7} + \sqrt{8 - 3\sqrt{7}} - \sqrt{8 + 3\sqrt{7}} = 0$
$\implies x\sqrt{7} = \sqrt{8 + 3\sqrt{7}} - \sqrt{8 - 3\sqrt{7}}$
$\implies 7x^2 = \left(8 + 3\sqrt{7}\right) + \left(8 - 3\sqrt{7}\right) - \underline{2\sqrt{\left(8+3\sqrt{7}\right)\cdot\left(8-3\sqrt{7}\right)}}$
$\implies 7x^2 = 16 - 2\cdot\sqrt{64-63} = 14$
$\implies x^2 = 2$
I know how to do all the steps up to the part where the underlined section comes into play, can someone please explain where does this come from. Thanks! 

Alt. hint:   show first that $\;\sqrt{8 \pm 3\sqrt{7}} = \dfrac{1}{\sqrt{2}}\left(3 \pm \sqrt{7}\right)\,$, then no squaring is needed.
