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,
$$[f(x^2 - x) + 1]f(x^3 + 1) = [f(x^4 - 2x^3 + x) + 1]f(x + 1) = (x - 1)x(x + 1)(x^6 - 3x^4 + 2x^3 + 1)$$

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.
