<s>[INST] Show $\frac{\sqrt{\sqrt[4]8-\sqrt{\sqrt2+1}\;}}{\sqrt{\sqrt[4]8+\sqrt{\sqrt2-1}\;} -\sqrt{\sqrt[4]8-\sqrt{\sqrt2-1}\;}}=\frac1{\sqrt2}$ Days ago, I tried to demonstrate this equality, reducing radicals, multiplying by the conjugate of the denominator, etc. But, I did not reach anything similar to the right side.
$$
\frac{\sqrt{\sqrt[4]{8}-\sqrt{\sqrt{2}+1}\;}}{\sqrt{\sqrt[4]{8}+\sqrt{\sqrt{2}-1}\;} -\sqrt{\sqrt[4]{8}-\sqrt{\sqrt{2}-1}\;}}=\frac{1}{\sqrt{2}}
$$
Can you help?  I lack vision for this problem. You may have fantastic tricks that I have not heard about.
Good 2020!
 [/INST] Apply the denest formula $\sqrt{a-\sqrt c}=\sqrt{\frac{a+\sqrt{a^2-c}}2}-\sqrt{\frac{a-\sqrt{a^2-c}}2} $ to the numerator 
$$N=\sqrt{\sqrt[4]8-\sqrt{\sqrt2+1}}=\sqrt{\frac{\sqrt[4]8+\sqrt{\sqrt2-1}}2}-\sqrt{\frac{\sqrt[4]8-\sqrt{\sqrt2-1}}2}=\frac1{\sqrt2}D$$
where $D$ is the denominator.

Alternatively, evaluate
$$D^2 = \left(\sqrt{\sqrt[4]8+\sqrt{\sqrt2-1}}-\sqrt{\sqrt[4]8-\sqrt{\sqrt2-1}}\right)^2$$
$$=\sqrt[4]8+\sqrt{\sqrt2-1}+\sqrt[4]8-\sqrt{\sqrt2-1}-2\sqrt{\sqrt8-(\sqrt2-1)}$$
$$=2\sqrt[4]8-2\sqrt{\sqrt2+1}=2N^2$$
Thus, $\frac ND = \frac1{\sqrt2}$.
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