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?
 A: $$II\;\;\;\;4\lambda y^3-4\lambda x=0\iff\;\lambda=0\;\;or\;\;x=y^3$$
and you already have a possible relation between $\,x,y\,$ ...and note it can not be $\,\lambda =0\;$ (why?)
A: As noted in several comments, the second equation yields $x=y^{3}$. Hence, by the third equation, $$y^{12}=3y^{4} \implies y^{8}=3$$
Assuming $y$ is real, we get $y=\pm 3^{1/8}$ and $x=\pm3^{3/8}$. By the first equation, then, (taking the positive roots): $$1+4\lambda3^{9/8}-4\lambda3^{1/8}=0 \implies 8\lambda3^{1/8}=-1$$
So $\lambda=\frac{-1}{8}3^{-1/8}$. If we were to take the negative roots, $\lambda$ would be positive.
NB: Notice that none of $x,y,\lambda$ can be zero due to equation $1$.
A: Hint: It is obvious (see I eq.) that $\lambda \ne 0$. Then from II eq. we have $x=y^3$. Now write this in III eq. and find possible values for $y$ and then $x$ and $\lambda$.
A: $\lambda \neq 0$ 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.
