Solve $x+3y=4y^3,y+3z=4z^3 ,z+3x=4x^3$ in reals 
Find answers of this system of equations in reals$$
\left\{ 
\begin{array}{c}
x+3y=4y^3 \\ 
y+3z=4z^3  \\
z+3x=4x^3  
\end{array}
\right. 
$$

Things O have done: summing these 3 equations give $$4(x+y+z)=4(x^3+y^3+z^3)$$
$$x+y+z=x^3+y^3+z^3$$
I've also tried to show that $x,y,z$ should be between $1$ and $-1$(As $(-1,-1,-1)$ and  $(1,1,1)$ are answers of this system of equations) but I was not successful.   

Consider $f(x) = 4x^3 - 3x$. We have: $f'(x) = 12x^2 - 3 = 0 \iff x = \pm \dfrac{1}{2}$. Thus:
If $x \in \left(-\dfrac{1}{2}, \dfrac{1}{2}\right)$ then $f'(x) < 0$, and $f$ decreases. This means:
If $x > y > z$, then: $x = f(y) < f(z) = y$, a contradiction. And we can obtain a contradiction for any other inequalities of $x,y,z$ using the same argument. The same argument works for $x \leq -\dfrac{1}{2}$ or $x \geq \dfrac{1}{2}$ since then $f$ increases. 
For if $x > y = z$, then: $y = f(z) > f(x) = z$, contradiction again. Thus we must have:
$x = y = z$, and we have: $x = y = z = \pm 1$ and $0$
