From the partial sum formula, $a_n=S_n-S_{n-1}=6n^3-6(n-1{)}^3$ for $n\ge 2$ (and $a_1=6$ which also fits). Expanding:

$a_n=6\left[n^3-(n-1{)}^3\right]=6\left[3n^2-3n+1\right]$

Then $a_{k+1}-a_k=6\left[3(k+1{)}^2-3(k+1)+1\right]-6\left[3k^2-3k+1\right]=6\cdot 6k=36k$.

Hence ${\left(\frac{a_{k+1}-a_k}{36}\right)}^2=k^2$, and

${\sum }_{k=1}^6{k}^2=\frac{6\cdot 7\cdot 13}{6}=91$.