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

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- May 2nd 2011, 05:14 AMpoirotshowing 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 AMTheChaz
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 AMtopsquark