Using induction prove $n^3-n$ is divisible by 3 whenever n is a positive number. I am not sure if I am doing this right, but I have this:
There exists an integer $k$.
$2k =$ positive number
$(2k)^3 - 2k$ [*And this is where I get lost. How does one prove this?]
 A: Note that $n^3-n=n(n^2-1)=n(n+1)(n-1)$. Now we know that of any three consecutive numbers is at least one divisible by 3, so this gets your result. 
A: For $n=1$, the result is trivial, so let's use $n=2$ as our base case: $2^3-2 = 8-2 = 6 \equiv 0 \bmod 3$.
Now, to use induction, we assume that it is true for $n$, and we want to prove that this must imply that it is true for $n+1$.
Let $3 \mid n^3-n$.
Now, $$\begin{align*}
(n+1)^3-(n+1) &= n^3+3n^2+3n+1-(n+1) \\
 &= n^3+3n^2+2n \\
 &= \color{red}{n^3 -n} + \color{blue}{3n^2 +3n}.
\end{align*}$$
Now, by our induction assumption, the term in red is divisible by $3$, and the term in blue is obviously divisible by $3$.
Therefore, by induction, $n^3-n$ is divisible by 3.
A: Let the above be a proposition denoted by $P_n$.
Base case:
Let $n=1$,
$1^3-1=0$ and since $0$ is divisible by $3$, $P_n$ holds for $n=1$.
Inductive step:
Let the proposition be true for $n=k$ given by,
$k^3-k$ is divisible by $3$
Then it holds for $n=k+1$.
$(k+1)^3-(k+1)$
=$k^3-k+3(k^2+k)$ We see that $3(k^2+k)$ is divisible by $3$ and since we assumed that $k^3-k$ is divisible by $3$ therefore it is proved that $P_n$ holds $\forall n \in \mathbb N$
A: Below are a handful of possible approaches.
First, notice $\displaystyle \ \ n^3\!-n  = n(n^2\!-1) = (n\!+\!1)n(n\!-\!1)  = \color{#c00}6{ {n\!+\!1\choose 3}} \ $ is divisible by $\,\color{#c00}6$
Alternatively show $\,(n\!+\!1)n(n\!-\!1)\,$ is divisible by both $2$ and $3$. Generally a sequence of $\,k\,$ consecutive integers contains a multiple of $\,k\,$ (hint: let pigeonholes be remainders mod $\,k).$
More generally $\ 2p\mid n^p-n\ $ for odd primes $p,\,$ since $\ p\mid n^p-n\ $ by little Fermat, and it is divisible by $2$ since $\,n^p$ and $\,n\,$ have equal parity. Even more generally 
Theorem $ $ (Korselt's Pseudoprime Criterion) $\ $ For  $\rm\:1 < e,m\in \Bbb N\:$ we have
$$\rm \forall\, n\in\Bbb Z\!:\ m\mid n^e\!-n\ \iff\  m\ \ is\ \  squarefree,\ \  and \ \ p\!-\!1\mid e\!-\!1\ \, for\  all \ primes\ \ p\mid m$$
For a proof see this answer. Your case is $\,\rm\,e,m = 3,6\,$ and, indeed, $2\!-\!1,\,3\!-\!1\mid 3\!-\!1.$
A: Without using induction you can note that $$n\cdot n^2\equiv n\cdot 1 \pmod 3$$ (using the Little Fermat's theorem and $n\neq 3k$)
therefore $n^3-n\equiv 0 \pmod 3$
A: We could also play it modular
$$n\equiv 0,1,2\pmod{3}$$
$$-n\equiv 0,2,1\pmod{3}$$
$$n^3\equiv 0,1,2\pmod{3}$$
And we can see that we always have 
$$n^3-n\equiv 0\pmod{3}$$
