Question

show that n3-n is divisible by 3 for any positive integers n

Answer 1

Hello ! I hope it helps .

= n^3 -n
= n(n^2-1)
= n(n^2 -1^2)
= n(n+1)(n-1)

We know if sum of digits is divisible by 3 , no. is divisible by 3 .

= n+n+1+n-1
= 3n (divisible by 3)

Hence , the n^3-n is divisible by 3 for any positive integer n .