$x^4+x^2=\frac{11}{5}$ then what is the value of $\sqrt[3]{\frac{x+1}{x-1}}+\sqrt[3]{\frac{x-1}{x+1}}$? 
$x^4+x^2=\frac{11}{5}$ then what is the value of $\sqrt[3]{\frac{x+1}{x-1}}+\sqrt[3]{\frac{x-1}{x+1}}$ ?

My approaches which didn't yield anything
I tried manipulating $\sqrt[3]{\frac{x+1}{x-1}}+\sqrt[3]{\frac{x-1}{x+1}}$ 
Observations

*

*$\frac{x+1}{x-1}=\frac{(x+1)(x^3-x^2+2x-2)}{(x-1)(x^3-x^2+2x-2)}=\frac{x^4+x^2-2}{x^4-2x^3+3x^2-4x+2}=\frac{\frac{11}{5}-2}{\frac{11}{5}-2x^3+2x^2-4x+2}$ which is not giving any assistance to find the required value.

*let a=$\sqrt[3]{\frac{x+1}{x-1}}+\sqrt[3]{\frac{x-1}{x+1}}$
then we observe that
$a^3=2\frac{x^2+1}{x^2-1}+3a$ which again is not useful

how should we proceed then to get the required ?

Let $a = ...$ "which again is not useful", wrong here, it is.
Let $a = x^2$
$a^2 + a - \frac{11}{5} = 0$
$a = \frac{-1 \pm \sqrt{1 - 4 * 1 * \frac{-11}{5}}}{2}$
$a = \frac{-1 \pm \sqrt{\frac{49}{5}}}{2}$
$a = \frac{-1 \pm \frac{7\sqrt{5}}{5}}{2}$
$a = \frac{-5\pm {7\sqrt{5}}}{10}$
$x^2 = \frac{-5\pm {7\sqrt{5}}}{10}$
$x = \pm\sqrt{\frac{-5+{7\sqrt{5}}}{10}}$
Let $y$ = $^3\sqrt{\frac{x+1}{x-1}}+$ $^3\sqrt{\frac{x-1}{x+1}}$
From your observation, $y^3 = \frac{2x^2+2}{x^2-1} + 3y$
Sub the value found for $x$:
$y^3 - 3y - 14\sqrt{5}-32 = 0$
Use the cubic formula without the $bx^2$ term in there. See origins of cubic formula for more info.
Without the $b$ constant we get a simplified version of the formula.
$y =$ $^3\sqrt{\frac{-d}{2a} + \sqrt{(\frac{-d}{2a})^2 + (\frac{c}{3a})^3}} +$ $^3\sqrt{\frac{-d}{2a} - \sqrt{(\frac{-d}{2a})^2 + (\frac{c}{3a})^3}}$
Note how this resembles the original part of the question we were finding for.
Another way to solve it:
Let $f(y) = y^3 - 3y - 14\sqrt{5}-32$
Cubics at most have $3$ real roots (atleast $1$ real, and the other $2$ complex).
In their factored form, $(x - a)(x - b)(x - c)$, $a*b*c = d$ in $ax^3+bx^2+cx+d$. Therefore, our roots must multiply to $-14\sqrt{5}-32$, or $abc = -14\sqrt{5}-32$. But before that we must define how many roots (1, 2 or 3) are there.
Taking $\frac{dy}{dx} = 0$, there is a maximum at $(-1, 61.305)$ and minimum at $(1, 65.305)$. Notice how the curves are way below the horizontal axis, so the $f(y)$ has only 1 root, for $y > 0$.
This means the $f(y)$ can be factored into a form resembling $(y + a)(y^2 + cy + d)$, where is $-a$ is the root and $(y^2 + cy + d)$ is irreducible.
So, $ay^2 + cy^2 = 0$, as there is no squared term. Therefore, $c = -a$.
$ad  = -14\sqrt{5}-32$
$acy + dy = -3y$
If you simultaneously solve, you will get $a = - 2 - \sqrt{5}$. Therefore, $(y - 2 - \sqrt{5}) = 0$ and your root is $2 + \sqrt{5}$.
