How to solve $8x^3-6x+6xy^2=0=4y^3-4y+6xy^2$

Question

How can I solve this system? $$ \left\{\begin{matrix} 8x^3-6x+6xy^2=0\\ 4y^3-4y+6xy^2=0 \end{matrix}\right. $$ I do $$ \left\{\begin{matrix} 8x^3-6x+6xy^2=0\\ 4y^3-4y+6xy^2=0 \end{matrix}\right. \Rightarrow \left\{\begin{matrix} 2x(4x^2-3+3y^2)=0\\ 2y(2y^2-2+3xy)=0 \end{matrix}\right.$$ Then $x=0 \Rightarrow y=0,\pm 1$ and $y=0 \Rightarrow x=0,\pm \frac{\sqrt{3}}{2}$

How can I find the other points?

Answer 1

The resultant of $8\,{x}^{3}+6\,x{y}^{2}-6\,x$ and $6\,x{y}^{2}+4\,{y}^{3}-4\,y$ with respect to $x$ is $32 y^3 (y-1)^2 (y+1)^2 (43 y^2-16)$. You've already done the $y=0$ case. For $y=\pm1$ you get $x=0$. For $y = \pm 4/\sqrt{43}$ you get $x = \pm 9/(2 \sqrt{43})$.

Answer 2

HINT: for $x\ne 0$ we get from the first equation $$6y^2=6-8x^3$$ thus we have $$y^2=\frac{3-4x^2}{3}$$ and from the second equation you will get $$x$$ $$x=\frac{2-2y^2}{3y}$$ this can you set in the equation above