I'm trying to solve a system of two cubic equations with two variables x and y.
The original problem was to solve the equation $z^3=-4i \overline{z}$. I know how to solve it using polar form.
Now I want to solve it using Cartesian form, say $z=x+yi$.
Doing the algebra and simplifying I got the next system of equations: $$\displaystyle\left\{\begin{matrix}x^3-3xy^2+4y=0\\-y^3+3x^2y+4x=0\end{matrix}\right.$$
It is trivial that $\displaystyle (0,0)$ is a solution, but I couldn't find the other four.
The best I got is $(3x^2-y^2)(x^2-3y^2)=16$, but I don't how to continue.
Please help, thanks!