prove that n^3-n+3 is divisible by 3 (no MI)

We can use a F.2 method (factorization and the idea of consecutive numbers) to solve this.
n^3-n+3
=n(n^2-1)+3
=n(n+1)(n-1)+3
Since one of the number must be a multiple of 3, n(n+1)(n-1) must be a multiple of 3 (if n=1 then n(n+1)(n-1)=0). And if a multiple of 3 is added by 3, it is still a multiple of 3.

Sorry , i don't know what is MI , so i just give my solution.
Any number ,which is divisible by 3, +3 is also divisible by 3
foe eg. 6+3 =9
so i neglect the +3
and take out the common factor 'n' in the n^3 -n
that become n(n^2-1)------(1)
there are three possible outcome for n/3
1. the remainder is 1
2. the remainder is 2
3 divisible by 3
so i can rewrite n into 3x,3x+1,3x+2
sub this three into (1)
if n=3x
3x ((3x)^2-1)
since there is a 3, it is divisible by 3
if n=3x+1
(3x +1)((3x+1)^2-1)
(3x+1)(9x^2+6x+1-1)
(3x+1)(3(3x^2+2x))
(3x+1)(3x^2+2x)3
since there is a 3, it is divisible by 3
if n=3x+2
(3x +2)((3x+2)^2-1)
(3x+2)(9x^2+36x+4-1)
(3x+2)(3(3x^2+12x+1))
(3x+2)(3x^2+12x+1)3
since there is a 3, it is divisible by 3
so no matter what number is n , n(n^2-1) is divisible by 3
thus, n^3-n+3 is divisible by 3