Question

for any positive integers not,prove that ncube -n is divisible by 6

Answer 1

n³ - n

= n(n²-1) 

= n(n+1)(n-1) 

= (n-1)(n)(n+1) 

n³-n is the product of three consecutive numbers. It is obvious that at least one of three consecutive numbers is even and divisible by 2 .

Also, n(n+1)(n-1) is divisible by 3 because 

n+n+1+n-1 = 3n ( 3n is divisible by 3 :- Divisibility rule of three ) .

As the result of n³-n is divisible by both 2 ,3 it is divisible by 6 as prime factors of 6 are 2 ,3 .

Hence proved!