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 approach and deduction is absolutely fine
Just for the sake of completeness, here is an alternative prove
Known
$$(a+b+c)^3 = a^{3} + b^{3}  + c^{3} + 3 a^{2} b + 3 a^{2} c + 3 a b^{2} + 3 a c^{2} + 3 b^{2} c + 3 b c^{2} + 6 a b c $$
$$ = a^{3} + b^{3}  + c^{3} + 3f(a,b,c)$$
Let $a=k-1,b=k,c=k+1$,then, $a,b,c$ represents consecutive numbers
So we have
$$(a+b+c)^3=(k-1+k+k+1)^3=(3k)^3=27k^3$$
Thus
$$a^{3} + b^{3}  + c^{3} = (a+b+c)^3 -  3\cdot f(a,b,c) = 27k^3-3\cdot f(a,b,c)$$
$$\Rightarrow a^{3} + b^{3}  + c^{3}=3\cdot(9k^3-f(a,b,c))$$
