Answer :

Let a = n^{3} – n

⇒ a = n × (n^{2} -1)

⇒ a = n × (n-1) × (n + 1) [Using (a^{2} –b^{2}) = (a−b) × (a + b)]

**⇒** a = (n-1) × n × (n + 1)

We know that

I If a number is completely divisible by 2 and 3, then it is also divisible by 6.

II If the sum of digits of any number is divisible by 3, then it is also divisible by 3.

III. If one of the factors of any number is an even number, then it is also divisible by 2.

Since, a = (n-1) × n × (n + 1)

Sum of the digits = n−1 + n + n + 1 = 3n which is a multiple of 3, where n is any positive integer.

And (n-1) × n × (n + 1) will always be even, as one out of (n-1) or n or (n + 1) must be even.

Hence, by condition I the number n^{3}-n is always divisible by 6, where n is any

positive integer.

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