Q: Prove that 6 divides n^(3) - n whenever n is a nonnegative integer?

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What I have is this:

Basis Step:

If n = 0, then f(n)= n^3 - n = 0^(3) - 0 = 0 - 0 = 0.

So the basis step is true since its divisible by 6.

Inductive Hypothesis:

We assume true, if n = k, then f(k) = k^(3) - k.

We must prove that if n = k + 1, then f(k+1) = (k+1)^(3) – (k+1)

Inductive Step:

f(k+1) = (k+1)^(3) – (k+1)

= (k + 1)(k + 2)^(2) - (k + 1)

= (k+1)(k^(2) + 2k + 1) – (k+1)

= k^(3) + 2k^(2) + k + k^(2) + 2k + 1 – k – 1

= k^(3) + 3k^(2) + 3k + 1 – k – 1

= k^(3) + 3k^(2) + 3k – k

= (k^(3) - k) + (3k^(2) + 3k)

= (k^(3) - k) + 3(k^(2) + k)

= f(k) + 3(k^(2) + k)

Where I am stuck at is I know f(k) is divisible by 6 but the 3(k^(2) + k) is divisible by 3?