<s>[INST] Solve this $\;2\sqrt[4]{\frac{x^4}{3} + 4} = 1 + \sqrt {\frac{3}{2}} |y| \ldots$ Solve
$\begin{cases}
2\sqrt[4]{\frac{x^4}{3} + 4} = 1 + \sqrt {\frac{3}{2}} |y|\\[8pt]
2\sqrt[4]{\frac{y^4}{3} + 4} = 1 + \sqrt {\frac{3}{2}} |x|
\end{cases}$
 [/INST] Following Ron Gordon's suggestion, and noting that the if $(x,y)$ is a solution, then $(\pm x, \pm y)$ are all solutions, we need to solve:
$$2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}}x$$
Both sides to the fourth power:
$$\frac{16x^4}{3}+64=1+2\sqrt6x+9x^2+3\sqrt6x^3+\frac{9x^4}{4}\\$$
It turns out that this expression factors:
$$(x-\sqrt6)^2(37x^2+38\sqrt6x+126)=0$$
So that $x=\sqrt6$.  The quadratic has no real roots.
Answers are then $(\pm\sqrt6,\pm\sqrt6)$.
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