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!
Best Answer
$$\displaystyle\left\{\begin{matrix}x^3-3xy^2+4y=0\\-y^3+3x^2y+4x=0\end{matrix}\right.$$ To solve this system of equations first multiply the first equation with $x$ and second with $y$. Then subtract the second equation from the first.
We get $$x^4+y^4-6x^2y^2=0$$
which is same as $$(x^2-2 x y-y^2) (x^2+2 x y-y^2)=0$$ So now we have two cases: $x^2-2xy-y^2=0$ or $x^2+2xy-y^2=0$
From here it should be pretty easy.