Solve a system of two equations with cubic radicals

Question

Solve the following system of equations ($x,y \in \Bbb R$):

$$\begin{cases} (8x-13)y&=(x+1)\sqrt[3]{3y-2}-7x \\ (y-1)x^2+(8y+7)x&=y^2+12y+(x+1)\sqrt[3]{3y-2}. \end{cases}$$

I think this system of equations have no solution, but I can't prove it.

Answer 1

HINT: we have $$(8x-13)y+7x=(y-1)x^2+(8y+7)x-y^2-12$$ from here we get (after simplifying) $$y(y-1)=x^2(y-1)$$ can you proceed?

Answer 2

HINT$$8xy-13y+7x=yx^2-x^2+8xy+7x-y^2-12y=(x+1)\sqrt[3]{3y-2}$$It follows $$(x^2-y)(y-1)=0$$ $y=1\Rightarrow x=1$ so $(x,y)=(1,1)$ is a solution.

On the other hand $x=\pm y$ gives the following two equations in $y$

$$\begin{cases} 512y^6+1152y^5+861y^4+209y^3-3y^2+3y+2=0\\512y^6-3840y^5+9603y^4-8011y^3+15y^2-9y+2=0\end{cases}$$

Both equations being irreducible, I rather stop here.