<s>[INST] Apporoaches to solve the given algebraic expression If  $\displaystyle \ \  x^{4} \ +\ x^{2} \ =\ \frac{11}{5}$  then what is the value of the given expression
$$\displaystyle \left(\frac{x+1}{x-1}\right)^{\frac{1}{3}} +\ \left(\frac{x-1}{x+1}\right)^{\frac{1}{3}} \ =\ \ ?$$
My Try :
As I can find the value of $\displaystyle x$, from the given equation but it will be tedious I think !.
$$\displaystyle x^{4} \ +\ x^{2} \ =\ \frac{11}{5}$$
$$\displaystyle \Longrightarrow \ x^{2} +1/2 \ =\ \frac{7}{2\sqrt{5}}$$
$$\displaystyle \Longrightarrow \ x^2 \ =\  \ \frac{7-\sqrt{5}}{2\sqrt{5}}$$
Which is getting too much complicated to solve the expression by putting the value of $\displaystyle x$.
What could be the other way to solve the given expression?
 [/INST] Consider $A = (\frac{x+1}{x-1})^\frac{1}{3} + (\frac{x-1}{x+1})^\frac{1}{3}$ and $x = \tan u$. Since $x^4+x^2 = \frac{11}{5}$ then $\frac{\sin^2u}{(1-\sin^2u)^2}=\frac{11}{5}$ by solving a simple quadric equation we get that $\sin u = \pm \sqrt{\frac{27-7\sqrt5}{22}}$.
Therefore $A = (\frac{\tan u+1}{\tan u-1})^\frac{1}{3} + (\frac{\tan u+1}{\tan u-1})^\frac{1}{3}$ so from the tangent formula for adding two arc we get: $-A = \tan^\frac{1}{3} {(u+\frac{\pi}{4})}+ \cot^\frac{1}{3} {(u+\frac{\pi}{4})}$. By rasing both sides to 3 we get:$$3A-A^3=\pm\frac{1}{\sin(u+\frac{\pi}{4})\times\sqrt{1-\sin^2{(u+\frac{\pi}{4})}}}$$On the other hand we know: $\sin(u+\frac{\pi}{4})=\pm\frac{\sqrt {225\sqrt 5-280}}{11}$. By simplifications and solving a simple polynomial equation of third degree we can easily get the final answer.
I hope that it can help you. Thank you for your question.
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