2023 AMC 12A Problem 12

Group into pairs $(2k)^3-(2k-1)^3$ for $k=1,\ldots,9.$ Expanding, $$\begin{gathered} (2k)^3-(2k-1)^3\\ {}=12k^2-6k+1. \end{gathered}$$ Summing for $k=1$ to $9,$ with $\sum k^2=285$ and $\sum k=45,$ gives $$\begin{gathered} 12\cdot 285-6\cdot 45+9\\ {}=3420-270+9\\ {}=3159. \end{gathered}$$ Thus, the correct answer is D.