How can we prove that r^3-r is a multiple of 6 for all natural values of r?

Question

-without mathematical induction-

Answer 1

r^3 + 0r^2 - r = r(r^2 + 0r - 1) = r(r+1)(r-1)

 

 (r-1)*r*(r+1) = 0*1*2, 1*2*3, 2*3*4, 3*4*5, 4*5*6, ...

 

Fact: The product of two consecutive integers is divisible by 2 and the product of three consecutive integers is divisible by 3 (and 2).