Solve a system of two equations with cubic radicals

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.

• Hint: $(x+1)\sqrt[3]{3y-2}$ seems to appear twice. – John B Jan 31 '16 at 12:09
• Where did the problem come from? Is it a model of some system? Please give some context. – Joel Reyes Noche Jan 31 '16 at 12:09
• one solution is $$x=1,y=1$$ – Dr. Sonnhard Graubner Jan 31 '16 at 12:21

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?

• Yes, you're right. How do we solve the system of equations $$\begin{cases} (8x-13)x^2&=(x+1)t-7x \\ t^3+2&=3x^2. \end{cases}$$ (in this case $y=x^2,\ t=\sqrt[3]{3y-2}$) – kimtahe6 Jan 31 '16 at 13:26

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.

• @ Piquito: I think that in this case $y=x^2$ you should let $t=\sqrt[3]{3y-2}$. So, we have $$\begin{cases} (8x-13)x^2&=(x+1)t-7x \\ t^3+2&=3x^2. \end{cases}$$. – kimtahe6 Jan 31 '16 at 13:37
• I fear that leads to change in the two last equations $y$ by $x^2$. Anyway, you can try your possibility. Regards. – Piquito Jan 31 '16 at 13:40
• Ok, but How do you can solve a equation with $\deg f=5$? – kimtahe6 Feb 1 '16 at 5:05
• With degree $5$ you try first at all to see if it is reducible. If it is not the case, then either, you see Wolfram (I don't like this way) and lie in some way about your answer, or honestly you leave the problem, – Piquito Feb 1 '16 at 14:01