On the {\(q\)}-analogue of the sum of cubes (Q1773159)

The sum, \(\sum_{k=1}^{n} k^3\), of the first \(n\) consecutive cubes is equal to \({{n+1}\choose{2}}^2\). The author gives a \(q\)-analogue of this result, \[ \sum_{k=1}^{n}q^{2n-2k} \frac{(1-q^k)^2(1-q^{2k})}{(1-q)^2(1-q^2)}=\left[ \begin{matrix} n+1\\ 2 \end{matrix} \right]^2 \] where \(\left[ \begin{matrix} n\\ k \end{matrix} \right] = \frac{(1-q^{n-k+1})(1-q^{n-k+2})\cdots(1-q^n)}{(1-q)(1-q^2)\cdots (1-q^k)}\), which is a simpler \(q\)-analogue than that discovered by [\textit{K. C. Garrett} and \textit{K. Hummel} [A combinatorial proof of the sum of \(q\)-cubes, Electron. J. Comb. 11, Research paper R9 (2004; Zbl 1050.05012)].