Solve $x+3y=4y^3,y+3z=4z^3 ,z+3x=4x^3$ in reals

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 $\mathbb{C}$.

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 $0$