Question : Let $a_r$ denote the following $$(1+x+x^2)^n=\sum_{r=0}^{2n}a_rx^r$$ then prove the following
$$\sum_{r=0}^{n}(-1)^r\binom n r a_r = \begin{cases} 0 & n \ne 3k \text{ for all } k\in\mathbb{N} \\ (-1)^\frac{4n}{3} \dbinom n {\frac{n}{3}} & n = 3k \text{ for all } k\in\mathbb{N} \end{cases}$$
My Attempt:
$$\sum_{r=0}^{n}(-1)^r\binom n r (1+x+x^2)^n$$ where coeffiecient of $x^r$ is the answer
$$(1+x+x^2)^n\sum_{r=0}^{n}(-1)^r\binom n r=0$$ hence my answer comes out to be $0$ regardless of whether $n$ is divisible by 3 or not
another similar approach I could think of is
$$(1-x+x^2)=\sum_{r=0}^{2n}(-1)^r a_r x^r$$ hence using this we get the original problem to convert into $$(1-x+x^2)^n \sum_{r=0}^{n}\binom n r= (1-x+x^2)^n \cdot 2^n$$ where coefficient of again $x^r$ is the answer which comes out to be $(-1)^ra_r\cdot2^n $ this time. Where am I going wrong?
