To check the divisibility of $n^3 - n$ by 6, we first factor the given expression. Taking $n$ as a common factor, we get:

$$n^3 - n = n(n^2 - 1)$$

We can further factor $n^2 - 1$ as $(n - 1)(n + 1)$. Thus, the complete factorization is:

$$n^3 - n = n(n - 1)(n + 1)$$

The factors of the expression are $(n - 1)$, $n$, and $(n + 1)$. As noted, any three consecutive numbers can be taken as $(n - 1)$, $n$ and $(n + 1)$. Therefore, the expression $n^3 - n$ represents the product of three consecutive natural numbers.

Now, we check for divisibility by 2. Among any three consecutive natural numbers, at least one number must be an even number. Therefore, the product of these three consecutive numbers, $n(n - 1)(n + 1)$, will always be divisible by 2. Next, we check for divisibility by 3. Among any three consecutive natural numbers, exactly one number must be a multiple of 3. Therefore, the product $n(n - 1)(n + 1)$ will always be divisible by 3. Since the expression $n(n - 1)(n + 1)$ is divisible by both 2 and 3, and these are co-prime numbers, the product must be divisible by their product, which is $2 \times 3 = 6$. Hence, $n^3 - n$ is always divisible by 6 for all natural numbers $n$.