How do I prove that \(n^3-n\) is always divisible with 6? I sort of see the solution since \(n(n-1)(n+1)\) will always have a part that is a 2 and a part that is 3. But I can't formulate it mathematically.

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3(2n) is always divisible by 6 ... not that it helps

the proof might be induction tho

Yes, induction is the best way.

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