# 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*}

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Notice that for every solution $(x, y)$, $(-x, -y)$ is also a solution.
$$(-1 + x y) (-2 + x^2 + y^2) = 0$$
Now solve for $y$ and substitute into the other equation, and solve for $x$. Positive solutions resulting from this are $x=y=1$ and $x = 2 y = 2 \sqrt{\frac{2}{5}}$.