Adopting the notation $[k] = x^k - x^{-k} $ and $[k]! = [2][3]...[k]$ (note that $[1]$ is omitted), and letting $m,n$ be two integers greater than $1$ such that $n>m$ and $gcd(m,n)=1$, would it be possible to write $$ \frac{1}{[m+n]}\binom{m+n}{m}_q = \frac{[m+n-1]!}{[m]![n]!} = \frac{[n+1][n+2]...[n+m-1]}{[m]!} = \frac{[n+1][n+2]...[n+m-1]}{[2][3]...[m]} $$ in the form: $$ \sum_{i=0}^k c_i (x^i+x^{-i}) $$
I know that if $l=mn$, then $\frac{[l]}{[n]} = x^{(l-1)n} + x^{(l-3)n} + ... + x^{-(l-1)n}$, but other than that I'm pretty much stuck.
Answers for some small $m,n$ are: $$(2,3): x^2 + x^{-2}$$ $$(2,5): x^4 + 1 + x^{-4}$$ $$(3,4): x^6 + x^2 + 1 + x^{-2} + x^{-6}$$ $$(3,5): x^8 + x^4 + x^2 + 1 + x^{-2} + x^{-4} + x^{-8}$$ $$(4,5): x^{12} + x^8 + x^6 + 2x^4 + x^2 + 2 + x^{-2} + 2x^{-4} + x^{-6} + x^{-8} + x^{-12}$$
Edit: The fraction should be some product of cyclotomic polynomials I think, because: $$x^k-x^{-k} = x^{-k}(x^{2k}-1)=x^{-k}\Pi_{d|2k} \Phi_d(x)$$