Proof that the sum of the cubes of any three consecutive positive integers is divisible by three. So this question has less to do about the proof itself and more to do about whether my chosen method of proof is evidence enough. It can actually be shown by the Principle of Mathematical Induction that the sum of the cubes of any three consecutive positive integers is divisible by 9, but this is not what I intend to show and not what the author is asking. I believe that the PMI is not the authors intended path for the reader, hence why they asked to prove divisibility by 3. So I did a proof without using the PMI. But is it enough? 
It's from Beachy-Blair's Abstract Algebra Section 1.1 Problem 21. This is not for homework, I took Abstract Algebra as an undergraduate. I was just going through some problems that I have yet to solve from the textbook for pleasure.  
Question: Prove that the sum of the cubes of any three consecutive positive integers is divisible by 3.
So here's my proof: 
Let a $\in$ $\mathbb{Z}^+$
Define \begin{equation} S(x) = x^3 + (x+1)^3 + (x+2)^3 \end{equation}
So, 
\begin{equation}S(a) = a^3 + (a+1)^3 + (a+2)^3\end{equation}
\begin{equation}S(a) = a^3 + (a^3 + 3a^2 + 3a + 1) + (a^3 +6a^2 + 12a +8) \end{equation}
\begin{equation}S(a) = 3a^3 + 9a^2 + 15a + 9 \end{equation}
\begin{equation}S(a) = 3(a^3 + 3a^2 + 5a + 3) \end{equation}
Hence, $3 \mid S(a)$.
QED

Your solution is fine, provided you intended to prove that the sum is divisible by $3$.  
If you intended to prove divisibility by $9$, then you've got more work to do!
If you're familiar with working $\pmod 3$, note @Math Gems comment/answer/alternative. (Though to be honest, I would have proceeded as did you, totally overlooking the value of Math Gems approach.)
