Find all the solutions of this equation:$x+3y=4y^3 \ , y+3z=4z^3\ , z+3x=4x^3 $ in reals. Find all the solutions of this equation
$x+3y=4y^3 \ , y+3z=4z^3\ , z+3x=4x^3 $
My attempt:In the hint of the question it was written that show $x,y,z \in [-1,1]$ then add all equations and conclude $(x,y,z)=(-1,-1,-1),(0,0,0),(1,1,1)$ are the solutions.I can show that $x,y,z \in [-1,1]$like below:
By symmetry consider $x \ge y \ge z$ then if $x>1$ we have:
$4x^3-4x>0 \Rightarrow 4x^3-3x>x \Rightarrow z>x$
which is wrong.Now consider $x<-1$ we have:
$x<-1 \Rightarrow y<-1 \Rightarrow 4y^3-4y<0 \Rightarrow 4y^3-3y<y \Rightarrow x<y$
By the same way we can get $x,y,z \in [-1,1]$.By adding the equations we can get:
$x+y+z=x^3+y^3+z^3$
But what should I do now?

Brute force. If we use $y=4z^3-3z$ and $z=4x^3-3x$ in $x=4y^3-3y$, we obtain a huge polynomial of $27$th degree in $x$ which can be factorized (by some not-human algebraic manipulator) as
$$4x(x-1)(x+1)(8x^3+4x^2-4x-1)(8x^3-4x^2-4x+1)\\(64x^6-112x^4+56x^2-7)(64x^6-32x^5-80x^4+32x^3+24x^2-6x-1)\\(64x^6+32x^5-80x^4-32x^3+24x^2+6x-1).$$
If we are interested in the integers solutions then we have only $x=0$, $x=1$ and $x=-1$ because the other factors do not have integer roots (check each constant term). 
Now $x\in\{-1,0,1\}$ implies that $z=4x^3-3x=x$, $y=4z^3-3z=x$ and therefore the integer solutions of the system are: $(0,0,0)$, $(1,1,1)$ and $(-1,-1,-1)$.
