<s>[INST] 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.   
 [/INST] Suppose that we had $x>1$.  Then, since $4x^3-3x > x$, we have $z>x$, and similarly $y>z$, $x>y$, contradiction.  By symmetry, we conclude that $x,y,z\in [-1,1]$.
So there exist $\alpha,\beta,\gamma \in [0,\pi]$ with $x=\cos \alpha$, $y=\cos \beta$, $z=\cos \gamma$.  By the formula for $\cos 3\alpha$, we can rewrite the system of equations as:
$$
\left\{ 
\begin{array}{c}
\alpha \equiv \pm 3 \beta\ (\operatorname{mod}\ 2\pi) \\ 
\beta = \pm 3 \gamma\ (\operatorname{mod}\ 2\pi)  \\
\gamma = \pm 3\alpha\ (\operatorname{mod}\ 2\pi)  
\end{array}
\right. 
$$
So we have $\pm 27\alpha \equiv \alpha$, so either $26\alpha \equiv 0$ or $28\alpha \equiv 0$.  We find that $\alpha = \pi k / 13$ or $\alpha = \pi k / 14$ for some nonnegative integer $k$.
This gives $27$ solutions,* $x=\cos{\frac{\pi k}{13}}$ for $0\leq k\leq 13$, and $x=\cos{\frac{\pi k}{14}}$ for $1\leq k\leq 13$.
For example, one solution is $(\cos\frac{\pi}{14},\cos\frac{9\pi}{14},\cos\frac{3\pi}{14})$.
* This is exactly the number we expect, up to multiplicity, when intersecting three cubic surfaces; in particular there are no more solutions over $\mathbb{C}$.
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