What is the sum of coefficients in a polynomial expansion?

[anemone]

Let $S=(x^3-x^2+x+3)^{2015}=a_0+a_1x+a_2x^2+a_3x^3+\cdots+a_{6044}x^{6044}+a_{6045}x^{6045}$, where $a_0,\,a_1,\,a_2,\cdots,a_{2015}$ are all integers.

Evaluate $a_0+a_2+a_4+\cdots+a_{6042}+a_{6044}$.

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[anemone]

Congratulations to the following members for their correct solutions::)

1. lfdahl
2. laura123

Here's lfdahl's solution:

Spoiler

Given:
\[S(x) = \left ( x^3-x^2+x+3 \right )^{2015}= a_0+a_1x+a_2x^2+...+a_{6044}x^{6044}+a_{6045}x^{6045}\]

Obviously: $S(-1) = 0$

Let $S_0 = a_0+a_2+a_4+...+a_{6044}$ and $S_1 = a_1+a_3+a_5+...+a_{6045}$

Then:

\[S(-1) = S_0-S_1=0 \: \: \: or \: \: \: S_0=S_1\]

\[S(1) = 4^{2015}= S_0+S_1=2S_0\: \: \: or \: \: \: S_0=\frac{1}{2}\cdot 4^{2015} = 2^{4029}\]

Here's laura123's solution:

Spoiler

Let be $f(x)=(x^3-x^2+x+3)^{2015}$ then $f(1)=(1^3-1^2+1+3)^{2015}=4^{2015}$ and $f(-1)=[(-1)^3-(-1)^2-1+3]^{2015}=0$.
Furthermore
$\dfrac{f(1)+f(-1)}{2}$=$\dfrac{(a_0+a_1+a_2+a_3+\cdots+a_{6044}+a_{6045})+(a_0-a_1+a_2-a_3+\cdots+a_{6044}-a_{6045})}{2}$=$a_0+a_2+a_4+\cdots+a_{6042}+a_{6044}$
then
$a_0+a_2+a_4+\cdots+a_{6042}+a_{6044}=\dfrac{f(1)+f(-1)}{2}=\dfrac{4^{2015}}{2}=\dfrac{2^{4030}}{2}=2^{4029}$.