Solving a nonlinear system of equations. 
Given that $x,y,z\in\mathbb R$, solve $$\begin{cases}6x^2-12x=y^3\\6y^2-12y=z^3\\6z^2-12z=x^3\end{cases}$$

I've tried adding the equalities but to no avail. I'd add what I've tried, but it'd be useless at best. I've tried using some cube identities, etc. but I seem to lack the general idea of a solution this problem could have.
P.S.: no calculus, logs or anything like that can be used. Just basic algebraic manipulations. Thanks.
 A: Let $f(x)=\sqrt[3]{6x^2-12x}$. Then solutions to this system of equations correspond to $f^3(x)=x$, that is, order-3 fixpoints of $f$. (Note that $f^3=f\circ f\circ f$ here and below.)
Numerically, I count no less than $7$ real solutions:
$$x\in\{-0.889,-0.527,0,0.0123,1.940,2.0000001,2.488\}.$$
These are in the cycles $-0.889\to2.488\to1.940$ and $-0.527\to2.0000001\to0.0123$, as well as the order 1 fixpoint $f(0)=0$.
I found these numbers by numerically finding the roots of $f^3(x)-x$ (which is an very complicated function). But we can bound the roots of this function, thus justifying the brute force solution of just looking for roots in that region.
First, note that any cycle must contain a nonpositive entry, because $f(x)<x$ for all $x>0$, Now $f(x)<x$ iff $6x^2-12x<x^3$ iff $x(x^2-6x+12)>0$, and $x^2-6x+12>0$ for all $x$ since the discriminant $\Delta=6^2-4\cdot12=-12$ is negative, so if the smallest element of the cycle is positive, $f(x)<x$ is a contradiction.
Since $6x^2-12x$ takes a minimum at $x=1$, $f(x)\ge f(1)=-\sqrt[3]6$, so this is our lower bound on any cycle in $f$. The upper bound is $f(-\sqrt[3]6)=3.46,$ because $f$ is decreasing for $x\le0$. Thus by inspecting the solutions to $f^3(x)=x$ in this region, be can be sure to find all solutions.
The final way of whittling down the solution space is to note that there is a bound on how wiggly $f^3(x)$ can be, given $f(x)$. Since $f$ is decreasing on $(-\infty,1]$ and increasing on $[1,\infty)$, $f^2$ will have $4$ regions of increasing/decreasing, and $f^3$ will have exactly $8$. (This happens because the minimum of $f$, $-\sqrt[3]6$, is less than the $x$ value for the minimum, that is, $1$.) Each one of these regions has a crossing with the line $y=x$ except the last one, where $f^3(x)<f^2(x)<f(x)<x$, as we have already observed. Thus, the $7$ solutions here are all of them.
I know of no closed form for any of them except for $0$. (Edit: Peter Sheldrick's solution gives these numbers as roots of a certain degree $24$ polynomial, which passes as a "closed form" in some circles. I'll let you be the judge.)
Here's a cobweb graph of the two cycles:

A: Symmetry dictates $x=y=z$ is a solution. Hence, we get $$x^3 - 6x^2+12x = 0\implies x(x^2-6x+12) = 0 \implies x = 0, x = 3 \pm i\sqrt3$$
