Consider \begin{ align*} G(N,M;\alpha,\beta,K,q) = \sum\limits_{ j\in\mathbb{ Z}}(-1)^jq^{ \frac{ 1}{ 2}Kj((\alpha+\beta)j+\alpha-\beta)}\left[\begin{ matrix}M+N\\N-Kj\end{ matrix}\right]_{ q}. \end{ align*} In this paper, we prove the non-negativity of coefficients of some cases of $G(N,M;\alpha,\beta,K,q)$. For instance, for non-negative integers $n$ and $t$, we prove that\\ \begin{ align*} G\left(n,n;\frac{ 4}{ 3}+\frac{ 3(3^t-1)}{ 2},\frac{ 5}{ 3}+\frac{ 3(3^t-1)}{ 2},3^{ t+1},q\right) \end{ align*} and \begin{ align*} G\left(n-\frac{ 3^t-1}{ 2},n+\frac{ 3^t+1}{ 2};\frac{ 8}{ 3}+2(3^t-1),\frac{ 4}{ 3}-(3^t-1),3^{ t+1},q\right)\\ \end{ align*} are polynomials in $q$ with non-negative coefficients. Using cubic positivity preserving transformations of Berkovich and Warnaar and some known formulae arising from Rogers-Szeg\"{ o} polynomials, we establish new identities such as\\ \begin{ align*} \sum\limits_{ 0\le 3j\le n}\dfrac{ (q^3;q^3)_{ n-j-1}(1-q^{ 2n})q^{ 3j^2}}{ (q;q)_{ n-3j}(q^6;q^6)_{ j}} = \sum\limits_{ j=-\infty}^{ \infty}(-1)^jq^{ 6j^2}{ 2n\brack n-3j}_q. \end{ align*}
Categories: math.NT math.CO