Solving system of equations: $x^3-3y^2x=-1$ and $3yx^2 -y^3=1$ $$x^3-3y^2x=-1$$
$$3yx^2 -y^3=1$$
This was the real part and imaginary part on a previous question I asked, instead of the system it was easier to just use polar coordinates to solve, but if this was just a system unrelated to the problem, how would one solve it? I seem to have trouble when the terms are like this. All I can see is that x and y are nonzero so we can divide by them.
How would you approach this system?
I have
$$x(x^2-3y^2)=-1$$
$$y(3x^2 -y^2)=1$$
 A: Eliminating $y$, we have
$$y^2=\frac{x^3+1}{3x}\ ,\quad y(3x^2-y^2)=1$$
and so
$$\frac{x^3+1}{3x}\Bigl(3x^2-\frac{x^3+1}{3x}\Bigr)^2=1\ .$$
Multiplying both sides by $27x^3$, the equation becomes a cubic in $x^3$.  The cubic happens to have one rational root, so all roots can easily be found.
A: HINT:
$$x^3+3x^2y-3xy^2-y^3=0$$
$$(x-y)(x^2+xy+y^2)+3xy(x-y)=0\iff(x-y)(x^2+xy+y^2+3xy)=0$$
Check both cases
A: *

*When $x = y$ the system reduces to $ x = y= t $ and $ 2t^3 = 1 $ hence the solution is $x = y= \dfrac{1}{\sqrt[3]2}$

*Suppose $x \neq y$ then the system reduces to  (after adding both equations) $$ x^3 - y^3 =  3y^2x - 3yx^2  $$


$$ \require{cancel} \cancel {(x - y)} (x^2 + xy + y^2) = - 3xy \cancel {(x - y)}  $$
$$ x^2 + 4xy  + y^2 = 0  $$
$$ [x + (2 - \sqrt 3) y] [x + (2 + \sqrt 3) y ] = 0  $$
Hence the set of solutions is; 
$$ (x, y  ) \in \{  (\frac{1}{\sqrt[3]2}, \frac{1}{\sqrt[3]2}) \} \bigcup \{ (x, y) \ | \ x \neq y \; \text{and} \; x = (\sqrt 3 - 2) y  \} \bigcup \{ (x, y) \ | \ x \neq y \; \text{and} \; x =  - (\sqrt 3 + 2) y  \}  $$
A: If you want to use complex numbers, here's an easier ad hoc way (I tried to describe how I found the solution step by step):
Note that, the terms involved arise in the expansion of $(x-y)^3$, but the signs are messed up. In particular, we need signs of the terms in the same equation to be same to use the formula.
$i^2=-1$ can be used to flip the signs. So, multiply first equation by $i^3$ and second by $i^2$. On adding the equations, LHS becomes $(ix+y)^3$. Solve using polar representation.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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Lets $\ds{3yx^{2} \equiv \cosh^{2}\pars{t}}$ and
$\ds{y^{3} \equiv \sinh^{2}\pars{t}}$ such that

\begin{align}
y&=\sinh^{2/3}\pars{t}\,,
\qquad
x={\cosh\pars{t} \over \root{3}y^{1/2}}
={\cosh\pars{t} \over \root{3}\sinh^{1/3}\pars{t}}
\end{align}

\begin{align}
-1&={\cosh^{3}\pars{t} \over 3\root{3}\sinh\pars{t}} - 3\sinh^{4/3}\pars{t}\,
{\cosh\pars{t} \over \root{3}\sinh^{1/3}\pars{t}}
\\[5mm]&={\root{3} \over 9}\,{\cosh^{3}\pars{t} \over \sinh\pars{t}} - \root{3}\sinh\pars{t}\cosh\pars{t}
\end{align}

\begin{align}
9\sinh\pars{t}&=\root{3}\bracks{9\sinh^{2}\pars{t} - \cosh^{2}\pars{t}}\cosh\pars{t}
\\[5mm]&=\root{3}\bracks{8\sinh^{2}\pars{t} - 1}\root{\sinh^{2}\pars{t} + 1}
\end{align}

With $\ds{z \equiv \sinh^{2}\pars{t}}$ we'll have:
\begin{align}
27z&=\pars{8z - 1}^{2}(z + 1)\ \imp\ 64z^{3} + 48z^{2} - 42z + 1=0\ \imp\
\left\{\begin{array}{rcl}
z_{0} & = & \phantom{-\,}\half
\\[2mm]
z_{1} & = & -\,{3\root{3} + 5 \over 8}
\\[2mm]
z_{2} & = & \phantom{-\,}{3\root{3} - 5 \over 8}
\end{array}\right.
\\[5mm]\mbox{and}&\qquad
x={\root{3} \over 3}\,{\root{1 + z} \over z^{1/6}}\,,\qquad
y=z^{1/3}
\end{align}
