Can I used polynomial congruence for prove 3 divide $n^3-n$? I'm not sure this method can work to prove  3|$n^3-n$? 
by let we have polynomial congruence 
$n^3-n\equiv 0(mod3)$
then if all residue class mod 3 are the roots of congruence 3|$n^3-n$
the residue for mod 3 are 0,1,2
then I plug each of them in congruence 
start by 0
$0-0\equiv 0(mod3)$ congruence 
then 1 
$1-1\equiv 0(mod3)$ congruence 
and 2
$8-2\equiv 0(mod3)$ congruence 
then concluded 3|$n^3-n$ 
Does it work ? 
 A: $n^3-n$ factors as $(n-1)n(n+1)$. For any integer $n$ it is easy to prove that one of the factors must be divisible by $3,$ so $n^3-n\equiv0\pmod3.$
A: Yes, it works. (You should state that it is $n$ you are taking the equivalence classes of-- you never actually stated that).  If $n \equiv 0, 1 ,2 \mod 3$ then $n^3 -n \equiv 0-0, 1-1, 8-2 \mod 3$ respectively.
You can make it a tad more wieldy you could say $n\equiv -1, 0, 1\mod 3$ to get $-1+1, 0-0, 1-1 \mod 3$ respectively.
And that is fine.  It does do everything a proof needs to do: it demonstrates irrefutably that the statement is true.
......
But 1) the text probably want a more intuitive, less brute force method and 2) for a more complex situation of a possibly high modulus such a proof may not be feasible or practical.
What they are getting at is to factor $n^3-n = n(n-1)(n+1)$.  Those, $n, n+1, n-1$ are three consecutive integers so one of them must be divisible by $3$.
.......
Consider instead $n^7-14n^5 + 49n^3-36n$ and proving it is divisible by $5040$ is going to be impractical be brute force ($5040$ cases to check!)  (Although a computer doing it would be an acceptable, albeit inelegant proof).
But factoring it as $(n-3)(n-2)(n-1)n(n+1)(n+2)(n+3)$ and noting it is a product seven consective integers so one of the terms is divisible by $7$, and least  one is by $5$.  Two are by $3$ and three are by $2$ with at least one by $4$.  So $7*5*3^2*2*2*4=5040$ divides it.
A: You can justify that the problem is reduced to remainders modulo $3$ like this:
Let $n=3q+r$ with $r\in\{0,1,2\}$ then 
$n^3-n=(3q+r)^3-(3q+r)=27q^3+27q^2r+9qr^2+r^3-3q-r\equiv r^3-r\pmod 3$
Then your method of evaluating the value for each $r$ is fine, and you get the conclusion it works for all integers.
