Multinomial Theorem:$$(a+b+c\dots)^n=\sum_{i+j+k\dots=n}{n!\over i!j!k!\dots}(a^ib^jc^k\dots)$$
Negative Binomial Theorem: $${1\over(a+b)^n}=\sum\binom{-n}{k}a^kb^{-n-k}=\sum(-1)^k\binom{n+k-1}{k}a^kb^{-n-k}$$
Im look for a way to combine the two
Specifically looking to expand ${1\over (1+x+x^2)^n}$but a general answer would also be good!
Edit: I'm looking for an approach without power series
COMMENT.-If you want to deduce from negative binomial the corresponding multinomial one, then you can do successively with $(a+X)^n$ where $X=b+c+...$, with (b+Y)^n with $Y=c+....$ and so on taking all in an single expression. The task is hard I think.
However for $1+x+x^2$ it could be less tedious in what involve powers but the coefficients? A way to simplify (going to binomial) is the simple identity $$1+x+x^2=\frac{x^3-1}{x-1}$$ so you have
$$\left(\frac{x^3-1}{x-1}\right)^{-n}=\frac{(x-1)^n}{(x^3-1)^n}$$