Might want to remember that

Let the three consecutive integers be $n - 1, n, n + 1.$ Then the sum of their cubes is

$(n - 1)^{3} + n^{3} + (n + 1)^{3} = (n^{3} - 3n^{2} + 3n - 1) + n^{3} + (n^{3} + 3n^{2} + 3n + 1) = 3n^{3} + 6n.$

Next, note that $3n^{3} + 6n - 9n = 3n^{3} - 3n = 3n(n^{2} - 1) = 3n(n - 1)(n + 1).$

Now for any three consecutive integers $n - 1, n, n + 1,$ precisely one of then is divisible by $3.$

Thus $3n(n - 1)(n + 1) = 3*3m = 9m$ for some integer $m,$ and so

$3n^{3} + 6n = 9m + 9n = 9(m + n)$ is divisible by $\boxed{9}$ for any integer $n.$

For sake of completeness, note that $2^{3} + 3^{3} + 4^{3} = 99,$ which is not divisible by any of $4,8$ or $12.$

Let the first integer be odd. Thus, the second is even and the third odd. We know that cubing an integer preserves parity. Hence, the sum of the cubes of three consecutive integers is odd if the first is odd. Then clearly, the sum of the cubes of three consecutive integers is not always even and will not always be divisible by any even number. Therefore, the answer is 9, since all other options are even.

Multiple choice for the win.