Factorization is an art (5)

Relevant wiki: Telescoping Series - Sum

$\begin{aligned} f(x) & = 4x^3-6x^2+4x-1 & \small \color{#3D99F6}{\text{Note that }(x-1)^4 = x^4 - 4x^3+6x^2-4x+1} \\ & = x^4 - (x-1)^4 \end{aligned}$

$\begin{aligned} \implies \sum_{-2016}^{2016} f(x) & = \sum_{-2016}^{2016} \left( x^4 - (x-1)^4 \right) \\ & = \sum_{-2016}^{2016} x^4 - \sum_{-2017}^{2015} x^4 \\ & = 2016^4 - 2017^4 \\ & = -\left(2017^4 - 2016^4\right) \\ & = - f(2017) \end{aligned}$

$\implies y = \boxed{2017}$

Here's my solution:

The key is to note that the function is (rotationally) symmetric about the point $(1/2,0)$ . We can see this because $f(1-x)=$ $4(1-x)^3-6(1-x)^2+4(1-x)-1=$ $-4x^3+6x^2-4x+1=$ $-f(x)$

So then, the terms in the series cancel out one by one, leaving only $f(-2016)=-f(1-(-2016))=-f (2017)$ .

$\large f(x)=4x^3-6x^2+4x-1$

$\large f(x)=(-x^4+4x^3-6x^2+4x-1) +x^4$

$\large f(x)=x^4-(x-1)^4$

$f(-2016)+f(-2015)+\dots+f(2015)+f(2016)$

$= \left(((-2016)^4-(-2017)^4)+((-2015)^4-(-2016)^4)+\dots+((-1)^4-(-2)^4)+((0)^4-(-1)^4)\right)$

$+\left( ((1)^4-(0)^4)+((2)^4-(1)^4)+\dots+((2015)^4-(2014)^4)+((2016)^4-(2015)^4)\right)$

$=-2017^4+2016^4$

$= -(2017^4-2016^4)$

$= -f(2017)$

$\Rightarrow y=2017$