OK, now I don't understand what you are trying to say.
There are n^2 results, each result takes O(n) to compute, so the naive algorithm is O(n^3). That's what you asked - why does the naive algorithm take O(n^3) - and that seems to answer your question.
Yes, clearly there is an overlap of some partial sums, and that's why a less naive algorithm is sub-O(n^3).
So, what are you asking, or what additional point are you making?
