<s>[INST] Solving a system of two cubic equations 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!
 [/INST] $$z^3=-4i\bar{z}\Rightarrow (-i)z^4=4|z|^2\Rightarrow |z|^4=4|z|^2$$
A trivial solution would be $|z|=0\Leftrightarrow z=0$. Assume $|z|\neq 0$ then 
$$|z|^2=4 \Rightarrow x^2+y^2=4$$
Expressing the main equation in cartesian coordinates:
$$(x+iy)^3=-4i(x-iy)\Rightarrow x^3+3ix^2y-3xy^2-iy^3=-4ix-4y$$
So one gets after equalizing the real and imaginary parts
$$x^3-3xy^2+4y=0$$
and 
$$3x^2y-y^3+4x=0$$
Adding the last two equations together yields
$$x^3-y^3+3xy(x-y)+4(x+y)=0\Rightarrow (x-y)(x^2+xy+y^2)+3xy(x-y)+4(x+y)=0\Rightarrow(x-y)(x^2+4xy+y^2)+4(x+y)=0$$ 
Using $x^2+y^2=4$ we get
$$(x-y)(1+xy)+(x+y)=0\Rightarrow x+x^2y-y-xy^2+x+y=0\Rightarrow x(2+xy-y^2)=0$$
So $x=0\Rightarrow y=0$ or $2+xy-y^2=0\Rightarrow 2+x\sqrt{4-x^2}-4+x^2=0$
The last result is equivalent
$$(2-x^2)^2=x^2(4-x^2)\Leftrightarrow 4-4x^2+x^4=4x^2-x^4\Leftrightarrow x^4-4x^2+2=0$$
This equation has four roots
$$x=\pm\sqrt{2-\sqrt{2}}$$
and 
$$x=\pm\sqrt{2+\sqrt{2}}$$
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