How to solve this difficult system of equations? $$1+4\lambda x^{3}-4\lambda y = 0$$
$$4\lambda y^{3}-4\lambda x = 0$$
$$x^{4}+y^{4}-4xy = 0$$
I can't deal with it. How to solve this?

$\lambda \neq 0$ since otherwise the first equation reads $1=0$. Also, $x \neq 0$ since then the second equation gives us $y=0$ which would lead to $1=0$ again in the first equation.
Now, the second equation gives us $x = y^{3}$. Plug into third equation to get $y^{12} = 3y^{4}$. Since $y\neq 0$, we get $y^8 = 3$. So $y$ can be any of the 8 roots of 3. $x$ is just $y^3$. Now $\lambda$ is easily solved from the first equation.
