Proof Verification: If $a\in \mathbb{Z}$, then $a^3≡a \pmod{3}$. So, I understand that congruence equation can be written as $a^3 -a = 3k$ ,$k\in \mathbb{Z}$. I also understand that this can be written as $a(a^2 - 1) = 3k$, which can be simplified further to show the multiplication of $3$ consecutive integers. 
$(a-1)a(a+1) = 3k$
I'm just not sure where to go from here. Can anyone help?

You simply need to realize that if you have $k$ consecutive integers one of them has to be divisible by $k$. 
The proof is almost too trivial to bother with.
If your consecutive numbers are $m, m+1,  m+2, m+3,...... , m+(k-1)$ and if $m \equiv i \pmod k$ then $m+1 \equiv i+1 \mod p$ and $m+2 \equiv i+2 \mod p$ and so on and then number $m + (k-i) \equiv i+(k-i) \equiv 0 \pmod k$.
(More generally, for any $j; 0 \le j < k-1$ then one of the consecutive integers is equivalent to $j \pmod k$.)
So if $(a-1), a, (a+1)$ are three consecutive numbers then one of them must be divisible by $3$.  
So the product of all of them is divisible by $3$ so $(a-1)a(a+1) \equiv 0 \pmod 3$.
....
Another less elegant but more direct one would be to note that either $a \equiv 0 \pmod 3$ or $a \equiv 1 \pmod 3$ or $a \equiv 2 \equiv -1 \pmod 3$.
So either $a^3 \equiv 0^3 = 0\equiv a \pmod 3$ or $a^3 \equiv 1^3 =1 \equiv a\pmod 3$ or $a^3 \equiv 2^3 = 8\equiv 2 \equiv a \pmod 3$.
Those are the only three options. 
....
Or if you really want a basic argument.  Let $a = 3k +i$ for some $k$ and $i = 0, 1, $ or $2$.
Then $a^3 = (3k + i)^3 = 27k^3 + 27k^2i + 9ki^2 + i^3 \equiv i^3 \pmod 3$.  And it's easy to verify $i^3 \equiv i \pmod 3$ if $i=0,1,2$.
