Solve the equaition ($\displaystyle x\in\mathbb{R}$):
$\displaystyle \frac 1{\left(1+\sqrt{1+\sqrt{x}}\right)^4}+\frac 1{\left(1-\sqrt{1+\sqrt{x}}\right)^4}+\frac 2{\left(1+\sqrt{1+\sqrt{x}}\right)^3}+\frac 2{\left(1-\sqrt{1+\sqrt{x}}\right)^3}=0$
Solve the equaition ($\displaystyle x\in\mathbb{R}$):
$\displaystyle \frac 1{\left(1+\sqrt{1+\sqrt{x}}\right)^4}+\frac 1{\left(1-\sqrt{1+\sqrt{x}}\right)^4}+\frac 2{\left(1+\sqrt{1+\sqrt{x}}\right)^3}+\frac 2{\left(1-\sqrt{1+\sqrt{x}}\right)^3}=0$
Hmm.. The solution is too long for me to write here, but let me express it. Let's call $\displaystyle \sqrt{1+\sqrt{x}} = a $
Then the equation is,
$\displaystyle \frac 1{\left(1+a\right)^4}+\frac 1{\left(1-a\right)^4}+\frac 2{\left(1+a\right)^3}+\frac 2{\left(1-a\right)^3}=0
$
What you must do here is to equalize the denominators by expanding all the fractions as they'll all have the denominator $\displaystyle \left(1+a\right)^4$$\displaystyle \left(1-a\right)^4 $
And then you can discard the common denominator, make some factorizations, simplify it as much as you can. Then this will give you a fourth degree equation which can be expressed as a quadratic equation. This will give you $\displaystyle a = \sqrt{1+\sqrt{\frac{8}{5}}}$ and you can get to $\displaystyle x = \frac{8}{5}$