Solve the system of radical equations

Question

Trinitrotoluen - Nguyen Thanh Nho - 2018

$x,y \in \mathbb{R}$ $$\begin{cases} (x-1)^2+(y+1)^2=2xy+1\\ \sqrt[3]{(3y-1)x^2-6y+1}-y= \frac{y^2-y(x-1)}{\sqrt[7]{x-1}} \\ \end{cases}$$

Correct?

Answer 1

There is an error after you reached,

$$3y^3+5y^2-5y=y^3$$

It should then reduce to,

$$y(2y^2+5y-5)=0$$

which yields the solutions $y=0, \frac14(-5\pm \sqrt{65})$, of which $y= 0$ is invalid.

Answer 2

With $$x=1+y$$ we get in the second equation $$(3y-1)(1+y)^2-6y+1=y^3$$ and this is $$y \left(2 y^2+5 y-5\right)=0$$ which is easy to solve.