TITLE: Let $f(x)=x^3-3x+1$.Find the number of distinct real roots of $f(f(f(x)))=3$
QUESTION [5 upvotes]: Let $f(x)=x^3-3x+1$.Find the number of distinct real roots of $f(f(f(x)))=3$
I have noticed that $f(x)=3$ has solutions $-1,-1,2$ But here how to find more roots ? in fact how to even say that they are roots of $f(f(f(x)))=3$?
If anyone has materials to share about these type of problems involving iterations, please do so.

REPLY [3 votes]: I write $a$ instead of $\{a\}$. The figure in lhs' answer shows that
$$f^{-1}(3)=\{-1,2\}\ .$$
One then obtains, by looking at the same figure,
$$f^{-1}(-1)=\{-2,1\},\quad f^{-1}(2)=\{\xi_1,\xi_2,\xi_3\}$$ with
$$\xi_1<-1,\quad -1<\xi_2<0,\quad 1<\xi_3<2\ .$$
Looking at the same figure again, we get
$$f^{-1}(-2)=\xi_4<-1,\quad f^{-1}(1)=\{\xi_5,0,\xi_6\},$$ $$ f^{-1}(\xi_1)=\xi_7,\quad f^{-1}(\xi_2)=\{\xi_8,\xi_9,\xi_{10}\},\quad f^{-1}(\xi_3)=\{\xi_{11},\xi_{12},\xi_{13}\}\ .$$
It follows that the set $f^{-3}(\{3\})\cap{\mathbb R}$ has $11$ elements. This is confirmed by the following Mathematica output: