$\displaystyle \frac 1{\left(1+\sqrt{1+\sqrt{x}}\right)^4}+\frac 1{\left(1-\sqrt{1+\sqrt{x}}\right)^4}+\frac 2{\left(1+\sqrt{1+\sqrt{x}}\right)^3}+\frac 2{\left(1-\sqrt{1+\sqrt{x}}\right)^3}=0$
Put $\displaystyle y=1+\sqrt{x}$, then your equation becomes:
$\displaystyle \frac 1{\left(1+\sqrt{y}\right)^4}+\frac 1{\left(1-\sqrt{y}\right)^4}+\frac 2{\left(1+\sqrt{y}\right)^3}+\frac 2{\left(1-\sqrt{y}\right)^3}=0$
$\displaystyle \frac {\left(1-\sqrt{y}\right)^4} {(1-y)^4} +
\frac {\left(1+\sqrt{y}\right)^4}{(1-y)^4}+
\frac {2\left(1-\sqrt{y}\right)^3}{(1-y)^3}+
\frac {2\left(1+\sqrt{y}\right)^3}{(1-y)^3}=0$
and since $\displaystyle y \ge 0$:
$\displaystyle \left(1-\sqrt{y}\right)^4 +
\left(1+\sqrt{y}\right)^4+
2\left(1-\sqrt{y}\right)^3(1-y)+
2\left(1+\sqrt{y}\right)^3(1-y)=0$
Now expand the powers and simplify and see what you have.
RonL
I used Wolfram Mathematica and got: $\displaystyle 14y^2+28y+6=0$
From this quadratic equation $\displaystyle y=\frac 17 \left(-7 \pm 2\sqrt{7}\right)\rightarrow x=\pm \frac 87 \left(-4 + \sqrt{7}\right)$
But it is not a solution. I can't find any mistakes in your solution (and I did it quite similarly) but the only solution is $\displaystyle \frac 85$ (by Mathematica ).
Let me check this out for you...
$\displaystyle 14y^2+28y+6=0$
$\displaystyle y^2+2y+\frac{3}{7}=0$
$\displaystyle y=\frac{-2 \pm \sqrt{4-4\frac{3}{7}}}{2}$
$\displaystyle y=\frac{-2 \pm \sqrt{\frac{28-12}{7}}}{2}$
$\displaystyle y=\frac{-2 \pm \sqrt{\frac{16}{7}}}{2}$
$\displaystyle y=\frac{-2 \pm 4\sqrt{\frac{1}{7}}}{2}$
$\displaystyle y=-1 \pm 2\sqrt{\frac{1}{7}}$
A solution of $\displaystyle y=1.6$ is impossible if
$\displaystyle 14y^2+28y+6=0$ is the quadratic equation. Notice that any positive number makes the equation greater than zero, and never less than or equal to zero.
Now I think you should get: $\displaystyle 10y^2-20y-6=0$
(which if I've got this right gives a root of $\displaystyle \approx 1.6$, which agrees with
the graphical sdolution)
(I must admit that I used "a tutors" substitution rather than my own as
it is easier to simplify - and then convert back as if I had made my substitution)
RonL