Question

for any positive integer n, prove that n³- n is divided by 6 ​

Answer 1

Answer: Leta= n^3-n= n.( n^2-1) = n.( n-1).( n+1)

sum of digits = n-1+ n+ n+1 =3n

= multiple of 3 , where n is the positive integer, then it is also divisible by 6

by this the number n^3- n is always divisible of 6.