Given that `n^3-n` can be factored as n(n+1)(n-1),

explain why `n^3-n` is divisible by 2 for any integer value of n.

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Since `n^3 - n` can be factored as *n*(*n - *1)(*n* + 1), for an integer value of *n*, it is a product of **three consecutive** **integers**. For example, 2, 3, 4, or 11, 12, 13.

Every other integer is even, or divisible by 2. If one lists ANY three consecutive integers, at least one of them has to be even.

If at least of of the integers in the produt is even (divisible by 2), then the whole product is divisible by 2 as well, because it contains a multiple of 2.

Therefore, the product of three consecutive integers, or *n*(*n* - 1)(*n + *1), is divisible by 2.

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