Yeah, Binomial King is here.

Let $f(z) = \left(3+z^{2015}+z^{2016}\right)^{2016}$ , then:

$\begin{aligned} f({\color{#3D99F6}x}) & = a_0 - \frac 12 a_1 - \frac 12 a_2 + a_3 - \frac 12 a_4 - \frac 12 a_5 + a_6 - \cdots \\ f({\color{#D61F06}z}) & = a_0 - \left(\frac 12 - i\frac {\sqrt 3}2\right) a_1 - \left(\frac 12 + i\frac {\sqrt 3}2\right) a_2 + a_3 - \left(\frac 12 - i\frac {\sqrt 3}2\right) a_4 - \left(\frac 12 + i\frac {\sqrt 3}2\right) a_5 + a_6 - \cdots \\ & = a_0 + a_1 e^{i\frac {2\pi}3} + a_2 e^{i\frac {4\pi}3} + a_3 e^{i\frac {6\pi}3} + a_4 e^{i\frac {8\pi}3} + a_5 e^{i\frac {10\pi}3} + a_6 e^{i\frac {12\pi}3} + \cdots \\ & = a_0 + a_1 \omega + a_2 \omega^2 + a_3 \omega^3 + a_4 \omega^4 + a_5 \omega^5 + a_6 \omega^6 + \cdots \end{aligned}$

$$

$\begin{aligned} \implies z & = \omega & \small \color{#3D99F6} \text{where }\omega \text{ is the third root of unity.} \\ f(x) & = \color{#3D99F6}\Re \left(f(\omega) \right) & \small \color{#3D99F6} \Re(z) \text{ is the real part of }z. \\ & = \Re \left[\left(3+\omega^{2015}+\omega^{2016}\right)^{2016}\right] & \small \color{#3D99F6} \text{Note that }\omega^3 = 1. \\ & = \Re \left[ \left(3+\omega^2+\omega \right)^{2016}\right] & \small \color{#3D99F6} \text{Note that }\omega^2 + \omega + 1 = 0. \\ & = \Re \left[ \left(3-1 \right)^{2016}\right] \\ & = \boxed{2^{2016}} \end{aligned}$