Functional equation involving $f(x^4)+f(x^2)+f(x)$ 
Find all increasing functions $f$ from positive reals to positive reals satisfying $f(x^4) + f(x^2) + f(x) = x^4 + x^2 + x$.

It's easy to show that $f(1)=1$, and I was also able to show that
$$f(x)-x = f(x^{(8^k)}) - x^{(8^k)}$$
for all integers $k$.
But where to go from here? Any hints would be much appreciated.

Let $g(x) = f(x)-x$. We obtain $g(x^4)+g(x^2)+g(x)=0$. Define $a_n = g(x^{2^n})$ for every $n\in\Bbb Z$ where $x\ne 1,x>0$. It follows that
$$
a_{n+2}+a_{n+1}+a_n=0,
$$ and from this 2nd order linear equation we get for some $A,B$,$$
a_n =\begin{cases}A,\quad n\equiv 0 \;\text{(mod }3)\\ B,\quad n\equiv 1 \;\text{(mod }3)\\-A-B,\quad n\equiv 2 \;\text{(mod }3)\end{cases}.
$$ 
It says that $(a_n)$ is a $3$-periodic sequence. Now we claim that $a_n = 0$ for every $n$. Since $f(x)=x+g(x)$ is increasing, it holds for each $k=0,1,2$ and all $n\equiv k\text{ (mod} \;3)$, $x>1$ that 
$$
x^{2^n}+a_k \le x^{2^{n+1}}+a_{k+1}\le x^{2^{n+2}}+a_{k+2}.
$$ Let $n\equiv k\text{ (mod} \;3)$ tend to $-\infty$ to obtain
$$
a_k\le a_{k+1}\le a_{k+2}
$$ for each $k=0,1,2$. This implies $a_0=a_1=a_2=0$ and thus $a_n = 0,\ \forall n$ follows for $x>1$. The case where $0<x<1$ can be treated similarly by noting that
$$
x^{2^n}+a_k \ge x^{2^{n+1}}+a_{k+1}\ge x^{2^{n+2}}+a_{k+2}.
$$
This proves $f(x)=x+g(x)=x$ for all $x>0$.
