Proof of for all integer $n \ge 2$, $n^3-n$ is divisible by 6 by mathematical induction.

Question

Prove the following statement by mathematical induction:
For all integer $n \ge 2$, $n^3-n$ is divisible by 6

My attempt:
[Proof] Let the given sentence p(n)
(1) $2^3-2$=6 is divisible by 6. p(2) is true.
(2) Suppose for all integer $k \ge 2$, p(k) is true. That is, mathematical hypothesis is $k^3-k$ is divisible by 6.

Then we must show that p(k+1) is true.

$(k+1)^3-(k+1) = (k^3 +3k^2+3k+1)-(k+1)$=$(k^3 -k)+(3k^2+3k)$

I have to get some multiple of 6 from $(3k^2+3k)$ to reach p(k+1) is true. How can I complete this proof?

Answer 1

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

Either $k$ or $k+1$ is even, so this is divisible by $3\cdot 2 = 6$.

Answer 2

Let's try this way- $n^3-n = n(n+1)(n-1)$

I make two statements - There exists one multiple of three in any three consecutive natural numbers

and There exists atleast one even number in any three consecutive numbers

If the above two statements make sense, Then, the given equation is divisible by $6$

Answer 3

If $k$ is odd then $3k^2 + 3k$ is even.

If $k$ is even then $3k^2 + 3k$ is even.