I came across the following identity in when I was reading my class lecture notes on Combinatorics. . It says using the $q$-binomial theorem, we can deduce that $$\prod_{j=1}^n (1-q^{3j-2})(1-q^{3j-1})=\sum_{j=-n}^n (-1)^jq^{j(3j+1)/2}\binom{2n}{n+j}_{q^3}.\tag{1}$$ Here $\binom{n}{k}_q$ denotes the $q$-binomial coefficient, which is defined by $$ \binom{n}{k}_q:=\frac{(1-q^n)(1-q^{n-1})\cdots (1-q^{n-k+1})}{(1-q^k)(1-q^{k-1})\cdots (1-q)}.$$ The $q$-binomial theorem is the following identity $$ (1+x)(1+xq)\cdots (1+xq^{n-1})=\sum_{j=0}^n \binom{n}{j}_q q^{j(j-1)/2}x^j.\tag{2}$$
I tried to use (2) to deduce (1) but cannot complete it. The steps are described as follows. By the $q$-binomial theorem, we have $$\prod_{j=1}^n(1-q^{3j-2})=\sum_{j=0}^n (-1)^j q^j q^{3j(j-1)/2}\binom{n}{j}_{q^3}$$ and $$ \prod_{j=1}^n(1-q^{3j-1})=\sum_{j=0}^n (-1)^j q^{2j} q^{3j(j-1)/2}\binom{n}{j}_{q^3}$$ Therefore, $$ \prod_{j=1}^n (1-q^{3j-2})(1-q^{3j-1})=\sum_{j=0}^{2n} (-1)^j\sum_{j_1+j_2=j} q^{j_1(3j_1-1)/2}\binom{n}{j_1}_{q^3}q^{j_2(3j_2+1)/2}\binom{n}{j_2}_{q^3}.$$ I am stuck here. I do not know how to transform the above expression into (1).
Using the hints given by @Somos, we have \begin{align*} \prod_{j=1}^{n}(1-q^{3j-2})(1-q^{3j-1})&=\prod_{j=1}^{n}(1-q^{3j-1})(-q^{3j-2})(1-q^{-3j+2}) \\&=(-1)^n q^{3n(n+1)/2-2n}\prod_{j=1}^{n}(1-q^{3j-1})\prod_{j=-n+1}^{0}(1-q^{3j-1}) \\&=(-1)^n q^{n(3n-1)/2}\prod_{j=-n+1}^{n}(1-q^{3j-1}) \\&=(-1)^n q^{n(3n-1)/2}\prod_{j=1}^{2n}(1-q^{3(j-n)-1}) \\&=(-1)^n q^{n(3n-1)/2}\sum_{j=0}^{2n}(-1)^j\binom{2n}{j}_{q^3}q^{3j(j-1)/2-(3n-2)j} \\&=\sum_{j=-n}^{n}(-1)^j\binom{2n}{n+j}_{q^3}q^{n(3n-1)/2+3(n+j)(n+j-1)/2-(3n-2)(n+j)}, \end{align*} giving the needed identity after simplification of the exponent.