By doing $(n+1)^3 - n^3$, I obtain $3n^2+3n+1$. Since the $+1$ prevents a multiple of 3 from forming, I understand the answer to this problem. However, what would be a more elegant manner to address why the +1 results in two consecutive terms never being a multiple of 3.
$3n^2 + 3n + 1 = 3(n^2 + n) + 1$.
Explanation: If $n$ is an integer, this will never be divisible by $3$ as $3(n^2 + n)$ is always divisible by $3$, but $1$ is not divisible by $3$. Thus the sum of these two is also never divisible by $3$.
In your proof, you can just mention the first line, but you have to add that this is never divisible by $3$ for all $n \in \mathbb Z$.