Show $\frac{\sqrt{\sqrt[4]8-\sqrt{\sqrt2+1}\;}}{\sqrt{\sqrt[4]8+\sqrt{\sqrt2-1}\;} -\sqrt{\sqrt[4]8-\sqrt{\sqrt2-1}\;}}=\frac1{\sqrt2}$

Question

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!

Answer 1

First square both sides and cross multiply to get

$$2\sqrt[4]{8}-2\sqrt{\sqrt 2+1}=2\sqrt[4]{8}-2\sqrt{2\sqrt[4]{4}-\sqrt 2 +1}$$

Notice that $$\sqrt[4]{4}=\sqrt 2$$ Thus the two sides are equal.

Answer 2

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.

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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, $\frac ND = \frac1{\sqrt2}$.

Answer 3

$$\frac{\sqrt{\sqrt[4]{8}-\sqrt{\sqrt{2}+1}\;}}{\sqrt{\sqrt[4]{8}+\sqrt{\sqrt{2}-1}\;} -\sqrt{\sqrt[4]{8}-\sqrt{\sqrt{2}-1}\;}}=\sqrt{\frac{\sqrt[4]{8}-\sqrt{\sqrt{2}+1}}{2\sqrt[4]8-2\sqrt{2\sqrt2-\sqrt2+1}}}=\frac{1}{\sqrt2}.$$