System of equations with radicals: $2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}y^2}$ and $2\sqrt[4]{\frac{y^4}{3}+4} = 1+\sqrt{\frac{3}{2}x^2}$ 
Solve the system of equations (in $\mathbb R$):
  $$\begin{matrix}
2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}y^2}
\\
2\sqrt[4]{\frac{y^4}{3}+4}
=
1+\sqrt{\frac{3}{2}x^2}
\end{matrix}.$$

This is from an older question, which was closed and deleted because of lack of context. (Here is link for users who can see deleted questions.) I will post my solution below - so I hope this time the question will not be closed for the lack of effort.
I found the system not too easy and somewhat interesting. (Of course I might have missed some straightforward way to the solution.) I'd be glad to see some other methods to solve it.

As you say, you will have solution corresponding to $x=\pm y$. Now, if you expand the last equation, you have (if $x \geq 0$),$$-\frac{37 x^4}{12}+3 \sqrt{6} x^3+9 x^2+2 \sqrt{6} x-63=0$$ To get rid of the $\sqrt{6}$'s, define $x=\sqrt{6} z$ and the equation becomes $$-111 z^4+108 z^3+54 z^2+12 z-63=0$$ By inspection $z=1$ is a solution. Making the long division let you with $$-111 z^3-3 z^2+51 z+63=0$$ where $z=1$ is solution again. Another division and arrive to $$-111 z^2-114 z-63=0$$ which does not show real solutions.
So, the real solutions are all possible combinations of $(\pm \sqrt{6},\pm \sqrt{6})$
