For any positive integer is n, prove that n³-n divisible by 6

1.	For any positive integer is n, prove that n³-n divisible by 6
Answer» Out of three (n\xa0– 1),\xa0n, (n\xa0+ 1) one must be even, so a is\xa0divisible\xa0by 2. 2. (n\xa0– 1) ,\xa0n, (n\xa0+ 1) are consecutive\xa0integers\xa0thus as\xa0proved\xa0a must be\xa0divisible\xa0by 3. Thus,\xa0n³\xa0–\xa0n\xa0is\xa0divisible by 6 for any positive integer n.

Discussion