CUbeSuM

We have the curious formula $\sum_{k=1}^{n}k^3=\left(\sum_{k=1}^{n}k\right)^2=\left(\frac{n(n+1)}{2}\right)^2.$ For $n=91$ this gives $\boxed{17522596}.$

We can prove this by induction, which is cheating a bit since we pull the formula "out of thin air."

Or we can use the standard approach: $k^4-(k-1)^4=4k^3-6k^2+4k-1$ Adding these up for $k=1...n$ , we find $n^4=4\sum_{k=1}^{n}k^3-6\sum_{k=1}^{n}k^2+4\sum_{k=1}^{n}k-n$ Solving for $\sum{k^3}$ and using the familiar formulas for $\sum{k^2}$ and $\sum{k}$ , we end up with $\sum_{k=1}^{n}k^3=\frac{n^4+n(n+1)(2n+1)-2n(n+1)+n}{4}=\left(\frac{n(n+1)}{2}\right)^2$ after a bit of routine algebra.

There must be a more direct way to derive this formula...