No number theory only

Let $f(ω) = ω^3-3ω \quad (1)$ then $\begin{cases}f(x)=x^3-3x=y \quad(2) \\ f(y)=y^3-3y=z \quad(3) \\ f(z)=z^3-3z=x \quad(4) \end{cases}$ and for $x \in (-\infty, -2]\cup [2,+\infty) \quad$ $f(x)$ is increasing thus WLOG: $x \le y \le z \quad (5)$ $\Rightarrow f(x) \le f(y) \le f(z) \\ \stackrel{(2)(3)(4)}{\Longrightarrow} y \le z \le x \quad (6)$ Hence considering $(5)$ and $(6)$ we conclude that $x=y=z$ which yields that $x^3-3x=x \\ \Rightarrow x(x-2)(x+2)=0 \\ \Rightarrow x = y = z = \{-2, 0, 2\}$ but $0 \notin (-\infty, -2]\cup [2,+\infty)$ therefore only $-2$ and $2$ are solutions.

Now we are left to see if there is any solution for $x \in (-2, 2)$ .

We can express $x$ as $x = 2 \cos w \quad (7)$ for $w \in (0, \pi)$

Using the identity $\boxed{\cos3\theta = 4cos^3\theta - 3\cos\theta} (*)$ (Check Chebychev Polynomials ) we have the following results:

$(2) \stackrel{(7)}{\Rightarrow} y = (2\cos w)^3-3(3\cos w) \Rightarrow y = 2(4\cos^3w-3\cos w) \stackrel{(*): \theta \rightarrow w}{\Longrightarrow} y = 2\cos3w\quad (8)$

$(3) \stackrel{(8)}{\Rightarrow} z = (2\cos 3w)^3-3(3\cos 3w) \Rightarrow z = 2(4\cos^33w-3\cos3w) \stackrel{(*): \theta \rightarrow 3w}{\Longrightarrow} z = 2\cos9w \quad (9)$

$(4) \stackrel{(9)}{\Rightarrow} x = (2\cos 9w)^3-3(3\cos 9w) \Rightarrow x = 2(4\cos^39w-3\cos9w) \stackrel{(*): \theta \rightarrow 9w}{\Longrightarrow} x = 2\cos27w \quad (10)$

Thus by $(7)$ and $(10)$ we got $x = 2\cos w = 2\cos 27w$ $\Rightarrow \cos 27w = \cos w \Rightarrow \begin{cases} 27w = w +2k\pi \\ 27w = -w + 2k\pi \end{cases} \Rightarrow \begin{cases} w =\dfrac{k\pi}{13} \quad(11)\\ w = \dfrac{k\pi}{14} \quad(12) \end{cases} \quad k \in \mathbb{Z}$ But recall that in $(7)$ we defined $0 < w < \pi \Rightarrow \begin{cases}(11) \Rightarrow 0 < k < 13 \\ (12) \Rightarrow 0 < k < 14 \end{cases}$

To sum up the solutions of $x$ are $\Big\{-2, \quad 2\cos \frac{k\pi}{13}(k=\{1,2,3,\ldots, 12\}), \quad 2\cos \frac{k\pi}{14}(k=\{1,2,3,\ldots, 13\}), \quad 2 \Big\}$