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For any positive integer n, prove that n...

For any positive integer `n`, prove that `n^3-n` is divisible by `6`.

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Let `a=n^(3) -n`
`rArr a=n(n^(2) -1)`
`rArr a=n(n-1) (n+1)`
Hence , n-1 ,n,n+1 are consecutive integes.
So out of these three consecutive integers at least one willl be divisible by 2.
`:.` a is divisible by 2.
Now sum of three consecutive integers =n-1+n+n+I =3n.
`:'` The sum of three consecutive integers is divisible by 3.
`:.` Exactly one of them must be divisible by 3.
Hence a is divisible by `2xx3 i.e., 6`
`rArr n^(3)-n` is divisible by6.
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