Factoring the Expression

We are given the expression $n^3 - n$, where $n$ is a natural number. We need to find the number that always divides this expression.

First, let's factor the expression:

• $n^3 - n = n(n^2 - 1)$
• Using the difference of squares formula ($a^2 - b^2 = (a-b)(a+b)$), we get $n^2 - 1 = (n-1)(n+1)$.
• So, $n^3 - n = n(n-1)(n+1)$.

Rearranging the terms, we have the expression as $(n-1) \times n \times (n+1)$.

Divisibility Properties of Consecutive Integers

The expression $(n-1) \times n \times (n+1)$ represents the product of three consecutive integers.

• Divisibility by 2: In any set of three consecutive integers, at least one integer must be even (divisible by 2).
• Divisibility by 3: In any set of three consecutive integers, exactly one integer must be divisible by 3.

Since the product contains a factor that is divisible by 2 and another factor that is divisible by 3, the entire product must be divisible by the least common multiple of 2 and 3, which is $2 \times 3 = 6$.

Conclusion

Therefore, $n^3 - n$ is always divisible by 6 for any natural number $n$.