for any positive integer n,p prove that n^3-n is divisible by 6
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n^3 - n = n(n^2-1)
= (n-1)n(n+1)
These are 3 consecutive integers
So one of them must be divisible by 3
Similarly , one/two of them must be divisible by 2
Thus , the whole term becomes divisible by 3*2=6
(There are theorems for the same but rather focus on quick answers than applying them)
= (n-1)n(n+1)
These are 3 consecutive integers
So one of them must be divisible by 3
Similarly , one/two of them must be divisible by 2
Thus , the whole term becomes divisible by 3*2=6
(There are theorems for the same but rather focus on quick answers than applying them)
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