Question

Prove that for any positive integers n, n^3- n is divisible by 6 ?

Answer 1

hope this helps you here is your ans

Answer 2

We can do this by mathamatical reasoningtr


The statement is true for n=0

Let assume k any +ve int and

K^3-k is true


Now for k+1 we have
( K+1)^3-(k+1)

=k^3+1+3k^2+3k-k-1
=k^3+3 k^2 +2k
=k(k^2+3k+2)
Which is a perfect square