Let the two consecutive numbers be n and (n+1). Difference of the cubes =(n+1)3−n3
=(n+1−n)[(n+1)2−n(n+1)+n2]
=n2+2n+1−n2−n+n2 =n2+n+1 =n(n+1)+1 Now, n(n+1) is always divisible by 2. So, n(n+1) is even. Thus, n(n+1)+1 is odd, which is never divisible by 2. Hence, the difference of two consecutive cubes is never divisible by 2. Trick: The cube of an odd number is always an odd number. The cube of an even number is always an even number and the difference between odd and even numbers is always odd. ∴ The difference of two consecutive cubes is never divisible by 2.