Proof:

To prove that n^3 - n is divisible by 6 for any positive integer n, we can use mathematical induction.

Base case:
Let's start with n = 1:

1^3 - 1 = 1 - 1 = 0

Since 0 is divisible by 6, the base case holds true.

Inductive step:
Assume that for some positive integer k, k^3 - k is divisible by 6. We need to prove that (k+1)^3 - (k+1) is also divisible by 6.

Expanding (k+1)^3 - (k+1):
(k+1)^3 - (k+1) = k^3 + 3k^2 + 3k + 1 - k - 1
= k^3 - k + 3k^2 + 3k

Using the assumption that k^3 - k is divisible by 6, we can write it as:
6m + 3k^2 + 3k (where m is an integer)

Factoring out 3:
3(2m + k^2 + k)

Now we need to show that (2m + k^2 + k) is divisible by 2. We can consider two cases:

Case 1: k is even
If k is even, then k^2 is also even. Let's say k = 2p, where p is an integer. Therefore, k^2 = (2p)^2 = 4p^2, which is divisible by 2. So, (2m + k^2 + k) is divisible by 2.

Case 2: k is odd
If k is odd, then k^2 is odd as well. Let's say k = 2p + 1, where p is an integer. Therefore, k^2 = (2p + 1)^2 = 4p^2 + 4p + 1 = 2(2p^2 + 2p) + 1, which is also divisible by 2. So, (2m + k^2 + k) is divisible by 2.

Since in both cases (2m + k^2 + k) is divisible by 2 and we already established that k^3 - k is divisible by 6, we can conclude that (k+1)^3 - (k+1) is divisible by 6.

Conclusion:
By proving the base case and the inductive step, we have shown that n^3 - n is divisible by 6 for any positive integer n.