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Prove True/False that n^3-n is Always Divisible By 6
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[QUOTE="CheesyPeeps, post: 5751934, member: 553192"] Okay, so I've factorised it and found that ##n^3-n## is always even and therefore always divisible by 2. In order for it to be divisible by 6, it must be divisible by 2 and 3, but I'm not sure how to go about proving that it's divisible by 3. EDIT: I looked up the divisibility rule for 3, and found it expressed as ##n(n+1)(n-1)## which is exactly what I have from factorising ##n^3-n##! [/QUOTE]
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Prove True/False that n^3-n is Always Divisible By 6
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