Answers

2015-07-05T18:42:43+05:30
N* (n2-1) is always divisible by 6
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2015-07-05T18:45:44+05:30
N^3-n=n(n+1)(n-1)
Case-1 where n is odd-
now if n=2a+1(a is any natural no.),2 divides (n+1)(n-1) again (n-1),n and (n+1) are three consecutive numbers.Hence 3 divides at least one of the three numbers-n,(n+1) and (n-1). So,3*2 or 6 divides n^3-n
Case-2 where n is even-
if n=2a(a is any natural number),2 divides n.So,2 divides n^3-n.Again, by the same logic 3 divides n^3-n.So,3*2 or 6 divides n^3-n when n is even.
Thus proved that 6 divides n^3-n .
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