Question 747469
not true:

n=1

n^3-n=0

Maybe you mean if n is 2 or greater?

{{{

n^3-n=(n)(n^2-1)=(n-1)(n)(n+1)
}}}


now for 

{{{n>=2}}}

this product will include 3 consecutive numbers and that means at least one of them will be even and exactly one will be a multiple of 3. Therefore each will be a multiple of 6.



:)