jmc

Question

The sum of the squares of three consecutive positive integers is 7805. What is the sum of the cubes of the three original integers?

Accepted Answer

If n is the middle of these integers, then we have (n-1)^2+n^2+(n+1)^2 = 3n^2+2 = 7805, or n^2 = 2601, meaning that n=51. Thus the sum of the cubes is 50^3+51^3+52^3 = 398259.