Find function $f(x)$ knowing that $[f(x^2 - x) + 1]f(x^3 + 1) = [f(x^4 - 2x^3 + x) + 1]f(x + 1)$ is satisfied for $\forall x \in \mathbb R$. 
Find function $f(x)$ knowing that $$\large [f(x^2 - x) + 1]f(x^3 + 1) = [f(x^4 - 2x^3 + x) + 1]f(x + 1)$$ is satisfied for $\forall x \in \mathbb R$ $(f(x) \ne c$ with $c$ being a constant$)$.

Okay, where to start. $f(x) = x$ doesn't work as $$(x^2 - x + 1)(x^3 + 1) - (x^4 - 2x^3 + x + 1)(x + 1) = 3(x - 1)x(x + 1)$$. Furthermore, if $f(x) = x^2 - x$ then $f(x^2 - x) = x^4 - 2x^3 + x$. However, at this case,
$$[f(x^2 - x) + 1]f(x^3 + 1) = [f(x^4 - 2x^3 + x) + 1]f(x + 1) = (x - 1)x(x + 1)(x^6 - 3x^4 + 2x^3 + 1)$$
 A: The function
$$f(x) = -\frac 1 2 x$$
satisfies the above property.
This is how I found it. First, notice that the equation $x^2 - x = x^4 - 2x^3 + x$ has solutions $x = -1, 0, 1, 2$. If you let $x = -1, 0, 1$ in the equality you get an identity, because $-1, 0, 1$ are also solutions of the equation $x^3 + 1 = x + 1$. If you let $x = 2$, though, you obtain
$$[f(2) + 1]f(9) = [f(2) + 1]f(3)$$
which means that either $f(2) = -1$ or $f(9) = f(3)$.
The simplest non-constant function such that $f(2) = -1$ is $f(x) = -\frac 1 2 x$. I checked if such function satisfied the property for all $x \in \mathbb R$, and indeed it does.
A: If we assume a solution of the form $f(x)=\dfrac{x}{a}$ then
$$ \left[\frac{1}{a}(x^2-x)+1\right]\cdot\left[\frac{1}{a}(x^3+1)\right]
=\left[\frac{1}{a}(x^4-2x^3+x)+1\right]\cdot\left[\frac{1}{a}(x+1)\right] $$
$$ (x^2-x+a)\cdot(x^3+1)=(x^4-2x^3+x+a)\cdot(x+1) $$
So either $x=-1$ or
$$ (x^2-x+a)\cdot(x^2-x+1)=x^4-2x^3+x+a $$
$$ x^4-2x^3+(a+2)x^2-(a+1)x+a=x^4-2x^3+x+a $$
$$ (a+2)x^2-(a+1)x=x $$
So $a=-2$ is the only solution.
Thus, as Luca Bressan found, not only is $f(x)=-\frac{1}{2}x$ a solution of the form $f(x)=cx$ it the only solution of that form.
ADDENDUM: By a similar analysis it can be shown that if $f(x)=\dfrac{1}{a}x+b$ then it must be the case that $a=-2$ and $b=0$.
