Answers

2014-05-16T18:21:16+05:30
If we prove that ( n 3 - n ) is divisible by 2 and 3,then it is also divisible by 6

Now ( n 3 - n ) = n ( n 2 - 1 ) = n ( n - 1 ) ( n + 1 ) 
Now whenever a number is divisible by 3 ,then the remainder obtained is either 0 or 1 or 2
∴ n=3p or 3p+1 or 3p+2 where p is some integer .
If n = 3p then n is divisible by 3
If n=3p+1 then n-1 = 3p+1-1 =3p is divisible by 3
If n = 3p+2 then n+1 = 3p+2+1 = 3p+3 is divisible by 3
Hence we can say that one of the numbers among n ,(n-1),(n-2) is always divisible by 3
Hence n(n-1)(n+1) is divisible by 3

Similarly when a number is divisible by 2 then the remainder obtained is either 0 or 1
∴ n=2q or 2q+1 where q is some integer
If n = 2q then n is divisible by 2
If n = 2q+1 then n-1 = 2q+1-1 = 2qis divisible by 2 and n+1 = 2q+2 is also divisible by 2
Hence we can say that one of the numbers amongn,(n-1),(n-2) is always divisible by 2
Hencen(n-1)(n+1) is divisible by 2

Now since n(n-1)(n+1) is divisible by 2 and 3 hence​n(n-1)(n+1) = n 3 - n ) ​is divisible by 6 6 as we know that if a number is divisible by 2 and 3 then it must be divisible by 6
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