prove n^{3}-n is always divisible by 6

*Since n*

^{3}

*– n*=

*n*(

*n*

^{2}– 1) =

*n*(

*n*–1)(

*n*+ 1).

Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.

∴

*n*= 3*p*or 3*p*+ 1 or 3*p*+ 2, where*p*is some integer.If

*n*= 3*p*, then*n*is divisible by 3.If

*n*= 3*p*+ 1, then*n*– 1 = 3*p*+ 1 –1 = 3*p*is divisible by 3.If

*n*= 3*p*+ 2, then*n*+ 1 = 3*p*+ 2 + 1 = 3*p*+ 3 = 3(*p*+ 1) is divisible by 3.So, we can say that one of the numbers among

*n*,*n*– 1 and*n*+ 1 is always divisible by 3.⇒

*n*(*n*– 1) (*n*+ 1) is divisible by 3.Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1.

∴

*n*= 2*q*or 2*q*+ 1, where*q*is some integer.If

*n*= 2*q*, then*n*is divisible by 2.If

*n*= 2*q*+ 1, then*n*– 1 = 2*q*+ 1 – 1 = 2*q*is divisible by 2 and*n*+ 1 = 2*q*+ 1 + 1 = 2*q*+ 2 = 2 (*q*+ 1) is divisible by 2.So, we can say that one of the numbers among

*n*,*n*– 1 and*n*+ 1 is always divisible by 2.⇒

*n*(*n*– 1) (*n*+ 1) is divisible by 2.Since, *n* (*n* – 1) (*n* + 1) is divisible by 2 and 3.

∴

*n*(*n*– 1) (*n*+ 1) =*n*^{3}*– n*is divisible by 6. (If a number is divisible by both 2 and 3, then it is divisible by 6)
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