# showing this is a multiple of 9

• May 2nd 2011, 05:14 AM
poirot
showing this is a multiple of 9
show (n-1)^3+n^3+(n+1)^3 is a multiple of 9

I expanded and have got 3n^3+6n so I need to show this is divisible by 9

3n^3+6n=9((1/3)n^3+(2/3)n)) but I can not gurantee the thing in the bracket is an integer
• May 2nd 2011, 05:27 AM
TheChaz
Quote:

Originally Posted by poirot
show (n-1)^3+n^3+(n+1)^3 is a multiple of 9

I expanded and have got 3n^3+6n so I need to show this is divisible by 9

3n^3+6n=9((1/3)n^3+(2/3)n)) but I can not gurantee the thing in the bracket is an integer

3n^3 + 6n = 3n(n^2 + 2)... I just factored this for ease in computation.
Now consider the remainder when n is divided by three.
n = 3k
n = 3k + 1 or
n = 3k + 2.
If n = 3k, then our expression 3n(n^2 + 2) = 3*(3k)*(9k^2 + 2) = 9k*something, is divisible by 9 (obviously).
If n = 3k + 1, then our expression 3n(n^2 + 2) = 3(3k + 1)*[(3k + 1)^2 + 2] = 3(3k + 1)*[3(3k^2 + 2k + 1)] = 9*something
If n = 3k + 2 is similar
• May 2nd 2011, 08:20 AM
topsquark
Quote:

Originally Posted by poirot
show (n-1)^3+n^3+(n+1)^3 is a multiple of 9

I expanded and have got 3n^3+6n so I need to show this is divisible by 9

3n^3+6n=9((1/3)n^3+(2/3)n)) but I can not gurantee the thing in the bracket is an integer

You have two good solutions already so I will just mention that induction is also a fairly easy way to do this as well.

-Dan