<s>[INST] Arithmetic sequence common difference involving cube or square! 
Evaluate $$\sum_{k = -3}^2 k^3$$

If I evaluate it using the expression above by putting the lower and upper limits it produces $-27$ for an answer, but when evaluated using the closed form expression of the geometric sequence where I put first term $a = -27$, $n = 3$ and common ratio $r = 3$, it produces the different result.  
I believe my common ratio $r = 3$ is incorrect. Can anyone tell me is $r$ value wrong? If yes, then how to calculate common ratio involving the cube or square etc.     
EDIT: Earlier I thought it was geometric sequence, but it is arithmetic sequence so I'm looking for d (common difference).  
EDIT I: As @Brian M. Scott pointed out, it is neither an arithmetic sequence nor geometric sequence. 
 [/INST] \begin{align}
S 
&= \sum_{k=-3}^2 k^3 \\
&= (-3)^3 + (-2)^3 + (-1)^3+ 0^3 + 1^3 + 2^3 \\
&= (-3)^3 - (2)^3 + 2^3 - 1 + 1 \\
&= -27
\end{align}
In general you could transform into a combination of sums of $j^k$, which is the topic of Faulhaber's formula.
\begin{align}
S 
&= \sum_{k=-3}^2 k^3 \\
&= \sum_{j=1}^6 (j-4)^3 \\
&= \sum_{j=1}^6 (j-4)(j^2 - 8j + 16) \\
&= \sum_{j=1}^6 (j^3 - 12j^2 + 48j-64) \\
&= \sum_{j=1}^6 j^3 - 12 \sum_{j=1}^6 j^2 + 48 \sum_{j=1}^6 j- 64 \sum_{j=1}^6 1 \\
&= \left. \left( \frac{n(n+1)}{2} \right)^2\right\vert_{n=6}
-12 \left. \frac{n(n+1)(2n+1)}{6} \right\vert_{n=6}
+48 \left.  \frac{n(n+1)}{2}\right\vert_{n=6} - 64 \cdot 6 \\
&= 441 - 1092 + 1008 - 384 \\
&= -27
\end{align}
which is overkill manually, but nice for a computer.
</s>