$$(1+x+x^2)^n = C_0 + C_1 x + C_2 x^2...C_n x^n$$
And $$(1-\frac 1x +\frac{1}{x^2})^n = C_0 -\frac{C_1}{x}+\frac{C_2}{x^2}...$$
So if we multiply, the two equations, the given series will be obtained by the coefficient of $x^{-1}$ on the LHS.
The LHS$$-(\frac{x^4+x^2 +1}{x^2})^n$$ The given answer is 0, but I don’t understand how $x^{-1} $ can have coefficient 0. If it is, how do I prove it?
$$ (1+x+x^2)(1-\frac 1x +\frac{1}{x^2}) = x^2 + 1 + \frac{1}{x^2} $$ so that $$ (1+x+x^2)^n(1-\frac 1x +\frac{1}{x^2})^n = (x^2 + 1 + \frac{1}{x^2})^n $$ contains only even powers of $x$. In particular, the coefficient of $x^{-1}$ is zero.