Prove True/False that n^3-n is Always Divisible By 6

[CheesyPeeps]

Homework Statement

For all natural numbers n, prove whether the following is true or false:
n³-n is always divisible by 6.

From SQA Advanced Higher Mathematics 2006 Exam Paper

Homework Equations

I can choose from the following types of proof:
Direct proof
Proof by contradiction
Proof by contrapositive
Proof by induction

The Attempt at a Solution

I know the statement is true, but proving it has been more difficult than I thought it would be!
I tried proof by induction, but got stuck with trying to prove true for n=k+1. I then tried proving the statement true for n=2k (even number) and n=2m+1 (odd number), but again, I didn't seem to be getting anywhere.
Am I along the right lines, or should I be trying something different?

 

[fresh_42]

Try to factor $n^3-n$ and see what you can say about the factors.

 

[andrewkirk]

Am I along the right lines

A simple direct proof is possible. Start by factorising $n^3-n$ into three integer factors.

EDIT: Now I am jinxed and will have to wait for somebody to say my name before I can speak aloud again.

 

[CheesyPeeps]

Try to factor $n^3-n$ and see what you can say about the factors.

Okay, so I've factorised it and found that $n^3-n$ is always even and therefore always divisible by 2. In order for it to be divisible by 6, it must be divisible by 2 and 3, but I'm not sure how to go about proving that it's divisible by 3.

EDIT: I looked up the divisibility rule for 3, and found it expressed as $n(n+1)(n-1)$ which is exactly what I have from factorising $n^3-n$!

 

[fresh_42]

What kind of numbers are $n-1\, , \,n\, , \,n+1$?