The sum of the cubes of 1, 2, 3 is 1+8+27=36 which is divisible by 9.

Now suppose that for some k: C(k)=k^3+(k+1)^3+(k+2)^3 is divisible by 9.

Then:

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.

RonL