Now suppose that for some k: C(k)=k^3+(k+1)^3+(k+2)^3 is divisible by 9.
C(k+1)=C(k) - k^3 + (k+3)^3 = C(k) + 3 (k^2*3) + 3 (k*3^2) + 3^3
..............=C(k) + 9 [k^2 + 3k + 3]
hence as by assumption C(k) is divisible by 9 so is C(k+1), and as we have
established that C(1) is divisible by 9, we have established by mathematical
induction that C(n) is divisible by 9 for all positive integers n.
This proves (as it is the same thing in different words) that the sum of the
cubes of three consecutive positive integers is divisible by 9.