Solve the system of equations $\begin{cases}x^3-3x=y \\ y^3-3y=z \\ z^3-3z=x \end{cases}$ 
Find the number of real solutions to the system of equations 
  $$\begin{cases}x^3-3x=y \\ y^3-3y=z \\ z^3-3z=x \end{cases}$$

Let $f(x) = x^3-3x$ then for $x\in \mathbb{R}-(-2,2)$ we have $x_1 < x_2 \Rightarrow f(x_1) < f(x_2)$. Thus
WLOG $x\le y\le z \qquad (1)$ $$\Rightarrow f(x) \le f(y) \le f(z) \\ \Rightarrow y\le z\le x\qquad (2)$$
Hence by (1) and (2) we conclude that $x=y=z$ which yields that $x=y=z= \{-2,0,2\}$
So there are 3 solutions, but this is not the correct answer.
 A: Let $f(x):=x^3-3x$ and $g(x):=f(f(f(x)))$.
We will look at possible values for variable $x$, values for $y$ and $z$ being computable from $x$ which are roots of the following equation 
$$h(x):=g(x)-x =0\ \ \ (*)$$
See curve at the bottom of this page.
We are going to show that the set of (strictly) positive roots of (*) is in two parts:
$$\left\{\begin{matrix}
x=2\cos(k \pi/13) \ \ \ k=1 \cdots 12\\
x=2\cos(k \pi/14) \ \ \ k=0 \cdots 14
\end{matrix}\right. \ \ \ \ (**)$$
Remark: one can check that the two sets in (**) have no common root. 
In this way, polynomial $h$ (which has degree 27) has $27$ real roots.
If we establish that all these $x$ are roots, we will have exhausted all of them.
Here is the proof. Let $x$ be a root of (*) in $[-2,2]$ (in fact, all of them are in this interval). Let:
$$x=2 \cos(a)  \ \ \ (1)$$
(a change of variable advised by @achille hui)
Multiplying trigonometry formula: $\cos(3a)=4\cos^3(a)-3\cos(a)$ by $2$, we obtain $2\cos(3a)=(2\cos(a))^3-3(2\cos(a))$, permitting to write $y$ as a function of $a$ in this way:
$$y=2 \cos(3a) \ \ $$
Along the same lines, we have: $ \ \ z=2 \cos(9a), \ \ \ $ and: 
$$x=2 \cos(27a) \ \ \ (2)$$
Let us write that (1) and (2) have to be compatible, i.e., $a$ must satisfy $\cos(27a)=\cos(a)$.
Two angles with the same cosine are either the same or are opposite (up to $k2 \pi$):
$$\left\{\begin{matrix}
27a=a+k 2 \pi & \ \text{giving} \ & a=k \pi/13\\
27a=-a+k 2 \pi & \ \text{giving} \ & a=k \pi/14
\end{matrix}\right. \ \ \ \ $$
which establishes (**).
The curve of $y=h(x)$:

