<s>[INST] Hard simultaneous equation problem $5x^2y-4xy^2+3y^3-2(x+y)=0$, $xy(x^2+y^2)+2=(x+y)^2$ How to solve this system of equations:
$$\begin{align*}
5x^2y-4xy^2+3y^3-2(x+y) &=0 \\ 
xy(x^2+y^2)+2 &=(x+y)^2
\end{align*}
$$
 [/INST] Notice that for every solution $(x, y)$, $(-x, -y)$ is also a solution.
The second equation admits factorization:
$$
  (-1 + x y) (-2 + x^2 + y^2) = 0 
$$
Now solve for $y$ and substitute into the other equation, and solve for $x$. Positive solutions resulting from this are $x=y=1$ and $x = 2 y = 2 \sqrt{\frac{2}{5}}$.
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