solve for real x and y: $x(x^2-3y^2)=2, y(3x^2-y^2)=11$ Solve for real x and y:
$x(x^2-3y^2)=2$
$y(3x^2-y^2)=11$
My attempt: I got $(x-y)^3=13$ but this doesn't always hold, I got a solution $(2,1)$.
How to proceed?
 A: The system can be written as
\begin{align*}
x^3-3xy^2&=2\\
3x^2y-y^3&=11
\end{align*}
Then
\begin{align*}
x^3-3xy^2+(3x^2y-y^3)i&=x^3+3ix^2y-3xy^2-iy^3=2+11i\\
(x+iy)^3&=2+11i\quad (\star)
\end{align*}
but
$$2+11i=8+12i-6-i=2^3+3(2^2)(i)+3(2)(i^2)+i^3=(2+i)^3$$
Now, there are three cubic roots of $2+11i$: $2+i,\;(2+i)\omega\;\text{and}\;(2+i)\omega^2$ where $\omega=\frac{-1+i\sqrt 3}{2}$.
From equation $(\star)$ we get
a) $x+iy=2+i\qquad \implies\qquad (x,y)=(2,1)$
b) $x+iy=(2+i)\omega=(2+i)\left(\frac{-1+i\sqrt 3}2\right)=-1-\frac{\sqrt 3}2+\left(-\frac12+\sqrt 3\right)i$, then $(x,y)=\left(-1-\frac{\sqrt 3}2,\;-\frac12+\sqrt 3\right)$
c) $x+iy=(2+i)\omega^2=(2+i)\left(\frac{-1-i\sqrt 3}2\right)=-1+\frac{\sqrt 3}2+\left(-\frac12-\sqrt 3\right)i$, then $(x,y)=\left(-1+\frac{\sqrt 3}2,\;-\frac12-\sqrt 3\right)$
A: NOTE: You made a mistake in getting $(x-y)^3=13$. Once you add the two equations, the left side is not $(x-y)^3$.
Idea:
\begin{align*}
\frac{x(x^2-3y^2)}{y(3x^2-y^2)}&=\frac{2}{11}.
\end{align*}
Let $y=mx$, then we have
\begin{align*}
\frac{(1-3m^2)}{m(3-m^2)}&=\frac{2}{11}.
\end{align*}
This gives
$$2m^3-33m^2-6m+11=0 \implies (2m-1)(m^2-16m-11)=0.$$
One of the solutions for this is $m=1/2$, now others can be found as well and they are $8 \pm 5\sqrt{3}$.
