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Solution
$$n^3-n$$ can be written as:
$$(n-1)n(n+1)$$ (where n is a positive integer)
i.e. product of three consecutive integers.
Hence for any number n=2 or >2 , product will have a factor of 6 in it.
When two numbers are prime in product, then third number will always be divisible by 6Â
Or product will always have a factor of $$3\times2$$ into it.
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