for any positive integer n,prove that n cube - n is divisible by 6

n³-n = n(n²-1)=n(n-1)(n+1)

 

For any number n, either n will be even or odd, if n will be odd then n+1 will be even, so the above term will have 2 as its factor and similarly, for any three consecutive numbers, one has to be divisble by 3. So any one of above is also divisible by 3. This means that there is 2 and 3 in the factor of above number, so it has to be divisble by 6.