Recurrence and polynomials?

Note that $f$ is strictly increasing on the real numbers. If $f (x)> x, f^{n+1}(x)> f^{n}(x)$ for all positive integers n, so $f^{2014}(x)> x$ . Similarly, $f (x)<x \implies f^{2014}(x)<x$ . Thus $f (x)=x \implies x^{3}-x+6=(x+2)((x-1)^{2}+2)=0$ . Hence $x=-2$ is the only solution.

The weird answer is to reduce the possibility of guessing.

Hi,

For convenience, let $f^{ 0 }\left( x \right) =x$ .

We can write $f\left( x \right) ={ x }^{ 3 }+6=\left( x+2 \right) g\left( x \right) -2$ where $g\left( x \right) ={ x }^{ 2 }-2x+4>1\quad \forall x$

Replacing $x$ with $f\left( x \right)$ yields

$f\left( f\left( x \right) \right) =\left( f\left( x \right) +2 \right) g\left( f\left( x \right) \right) -2=\left( x+2 \right) g\left( x \right) g\left( f\left( x \right) \right) -2$

$\Rightarrow f^{ n }\left( x \right) =\left( x+2 \right) \prod _{ m=0 }^{ n-1 }{ g\left( f^{ m }\left( x \right) \right) } -2,\quad n=1,2,3,...$

Now, we have all we need to solve the equation

$f^{ n }\left( x \right) =x\\ \Leftrightarrow \left( x+2 \right) \prod _{ m=0 }^{ n-1 }{ g\left( f^{ m }\left( x \right) \right) } =x+2\\ \Leftrightarrow \left( x+2 \right) \left( \prod _{ m=0 }^{ n-1 }{ g\left( f^{ m }\left( x \right) \right) } -1 \right) =0\\ \Leftrightarrow x=-2$