$2$-cycles of the map. Suppose that $f(p)=q$ and $f(q)=p$

Question

Considering the cubic map $x_{n+1}=f(x_n)$ where $$f(x_n)=rx_n-x_n^3$$

I am trying to find the $2$-cycles of the map. Suppose that $f(p)=q$ and $f(q)=p$

My textbook gives the answer but I am wondering about the first step about how this $f(p)=q$ and $f(q)=p$ is being applied.

So I know that $p \ne q $

and we let $$f(x)=rx-x^3$$

To find fix points we let $f(x)=x$, so

$$x=rx-x^3$$

Here is the part I do not understand and am looking for some clarification on,

$$1) f(p)=q, f(q)=p$$

$$2) x=r(rx-x^3)-(rx-x^3)^3$$

So how did they go from $1)$ to $2)$?

Answer 1

$2)$ follows from $1)$, since if $f(p)=q$ and $f(q)=p$, then when we let $f(x)=rx-x^3$, we also let $f(rx-x^3)=x$. So in equation $f(x_n)=rx_n-x_n^3$, write $rx-x^3$ instead of $x_n$ and you get $f(x)=r(rx-x^3)-(rx-x^3)^3=x$.