Calculate $\frac{a^8 + b^8}{a^8 - b^8} + \frac{a^8 - b^8}{a^8 + b^8}$ based on $\frac{a^2 + b^2}{a^2 - b^2} + \frac{a^2 - b^2}{a^2 + b^2}$.

Question

Given that $a$ and $b$ are distinct reals,

calculate $\dfrac{a^8 + b^8}{a^8 - b^8} + \dfrac{a^8 - b^8}{a^8 + b^8}$ based on $\dfrac{a^2 + b^2}{a^2 - b^2} + \dfrac{a^2 - b^2}{a^2 + b^2}$.

Well... this problem is adapted from a recent competition... And I can't solve it. Please help me.

Answer 1

Let $$x=\frac{a^{2}+b^{2}}{a^{2}-b^{2}}+\frac{a^{2}-b^{2}}{a^{2}+b^{2}}=2\frac{a^{4}+b^{4}}{a^{4}-b^{4}}.$$ Then $$\frac{a^{4}+b^{4}}{a^{4}-b^{4}}+\frac{a^{4}-b^{4}}{a^{4}+b^{4}}=\frac{x}{2}+\frac{2}{x}.$$ We can repeat this to get $$\frac{a^{8}+b^{8}}{a^{8}-b^{8}}+\frac{a^{8}-b^{8}}{a^{8}+b^{8}}=\frac{\frac{x}{2}+\frac{2}{x}}{2}+\frac{2}{\frac{x}{2}+\frac{2}{x}}=\frac{x}{4}+\frac{1}{x}+\frac{4x}{x^{2}+4}.$$