If $$\sum_{k=1}^{n} a_k = 6n^3$$, then $$\sum_{k=1}^{6} \left(\frac{a_{k+1} - a_k}{36}\right)^2$$ is equal to __________.

Let $$S_n$$ denote the partial sum $$S_n=\sum_{k=1}^{n}a_k$$. It is given that $$S_n=6n^3$$.

The general term of the sequence is obtained from consecutive partial sums: $$a_k=S_k-S_{k-1}$$ with $$S_0=0$$.

Compute $$a_k$$: $$a_k=6k^3-6(k-1)^3$$ $$=6\left[k^3-\left(k^3-3k^2+3k-1\right)\right]$$ $$=6\left(3k^2-3k+1\right)$$ $$=18k^2-18k+6.\;-(1)$$

Next, find the difference $$a_{k+1}-a_k$$. Using $$-(1)$$: $$a_{k+1}=18(k+1)^2-18(k+1)+6$$ $$=18(k^2+2k+1)-18k-18+6$$ $$=18k^2+36k+18-18k-18+6$$ $$=18k^2+18k+6.$$

Hence $$a_{k+1}-a_k=(18k^2+18k+6)-(18k^2-18k+6)$$ $$=36k.$$

Divide by $$36$$: $$\frac{a_{k+1}-a_k}{36}=k.$$

Square and sum from $$k=1$$ to $$6$$: $$\sum_{k=1}^{6}\left(\frac{a_{k+1}-a_k}{36}\right)^2=\sum_{k=1}^{6}k^2.$$

The standard formula for the sum of squares is $$\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}.$$ For $$n=6$$: $$\sum_{k=1}^{6}k^2=\frac{6\cdot7\cdot13}{6}=7\cdot13=91.$$

Therefore, the required value is $$91$$.