Mathematics question

Question

This is just a part of the solution of a problem.

p(n)=2n³–6n²+4n−3

=2n³–2n−6n²+6n−3

=2(n−1)n(n+1)−6n(n−1)−3

It is clear that 3 ∣ p(n) for each integer n.

Why is 3 ∣ p(n)?

Answer 1

The 2nd term of that final expression is -6(n)(n-1) is divisible by 3 because the factor 6 is divisible by 3.

The 3rd term is a constant (-3) which is also divisible by 3.

It is not so clear about the first term: 2(n)(n-1)(n+1). If you think about what those numbers could be, for example let n =7, then the term becomes 2(7)(6)(8). The sequence (n-1)(n)(n+1) is like counting by 3's. One of those factors has to be divisible by 3. If one of those factors is 0, the entire term is 0, which is also divisible by 3.