Proof $\frac{-\sqrt{2 - \sqrt{2 - \sqrt{2}}}}{\sqrt2} + \frac{\sqrt{2 + \sqrt2}}{\sqrt2 \sqrt{2 - \sqrt{2 - \sqrt2}}} =\sqrt{2-\sqrt{2+\sqrt2}}$ I want to prove the following equation
$$\frac{-\sqrt{2 - \sqrt{2 - \sqrt{2}}}}{\sqrt{2}} + \frac{\sqrt{2 + \sqrt{2}}}{\sqrt{2} \sqrt{2 - \sqrt{2 - \sqrt{2}}}}$$
I know this must be equal to $\sqrt{2-\sqrt{2+\sqrt{2}}}$, but I find it pretty hard to prove it manually, without requiring a calculator nor estimates (since the expression should be equal to one of the roots of the polinomyal $f(x)=x^8-8x^6+20x^4-16x^2+2$, and all of them are real and distinct).
Thanks in advance for your help.

$a=\sqrt{2-\sqrt{2-\sqrt{2}}}+\sqrt{2-\sqrt{2+\sqrt{2}}}$
$a^2=4+\sqrt{2}$
$a=\sqrt{2(2+\sqrt{2})}$
$b=\sqrt{2-\sqrt{2-\sqrt{2}}}*\sqrt{2-\sqrt{2+\sqrt{2}}}=\sqrt{4+\sqrt{2}-2(\sqrt{2-\sqrt{2-\sqrt{2}}}+\sqrt{2-\sqrt{2+\sqrt{2}}})}=\sqrt{4+\sqrt{2}-2(\sqrt{2(2+\sqrt{2})})} $
$-\sqrt{2-\sqrt{2-\sqrt{2}}}/\sqrt{2}+\sqrt{2+\sqrt{2}}/\left(\sqrt{2}\sqrt{2-\sqrt{2-\sqrt{2}}}\right)=\sqrt{2-\sqrt{2+\sqrt{2}}}?$
$-\sqrt{2-\sqrt{2-\sqrt{2}}}+\sqrt{2+\sqrt{2}}/\left(\sqrt{2-\sqrt{2-\sqrt{2}}}\right)=\sqrt{2(2-\sqrt{2+\sqrt{2}}})$
$\sqrt{2+\sqrt{2}}-(2-\sqrt{2-\sqrt{2}})=\sqrt{2(2-\sqrt{2+\sqrt{2}})(2-\sqrt{2-\sqrt{2}})}$
$\sqrt{2-\sqrt{2-\sqrt{2}}}+\sqrt{2-\sqrt{2+\sqrt{2}}}-2=\sqrt{4+\sqrt{2}-2(\sqrt{2(2+\sqrt{2})})}$
$\sqrt{2(2+\sqrt{2})}-2=\sqrt{2(4+\sqrt{2}-2(\sqrt{2(2+\sqrt{2})})}$
$\sqrt{2+\sqrt{2}}-\sqrt{2}=\sqrt{4+\sqrt{2}-2(\sqrt{2(2+\sqrt{2})})}$
Squared:
$4+\sqrt{2}-2(\sqrt{2(2+\sqrt{2})})=4+\sqrt{2}-2(\sqrt{2(2+\sqrt{2})})$
The proof is done.
