Question 990042: prove that n^3-n is always divided by 6
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! We can factor n^3 - n as:
n^3 - n = n(n^2-1) = n(n-1)(n+1)
So for every number n, we have the product of three consecutive integers.
For example:
3,4,5
6,7,8
12,13,14 etc.
In all these cases, there will be at least one even number, which is divisble by 2.
And in any trio of consecutive numbers, there will always be one number which is divisible by 3.
Thus the product of any three consecutive integers has factors of 2 and 3, which means it is divisible by 6.
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