Proof by induction that $3\mid n^3-n$ - confused about inductive step I have these notes, but I'm confused on what is happening at the inductive step. 
Theorem: 
$\forall n \in  \mathbb{N} 3 | (n^3-n) $
Inductive Step:
For $n \geq 0, show P(n) \Rightarrow P(n+1) is True$
Assume P(n) true i.e. $3 | (n^3-n) $
Examine: 
$(n+1)^3 - (n+1) $ = $n^3 + 3n^2 + 3n + 1 - (n+1)$
= $n^3 + 3n^2 + 2n  $
= $n^3 - n + 3n^2 + 3n$
I understand what we're trying to prove but 


*

*where does $(n+1)^3 - (n+1) $ come from? 

*What is happening here $n^3 - n + 3n^2 + 3n$, how did we get there from the previous point? I assume $n^3 + 3n^2 + 2n  $  is not obvious if it was divisible by 3, thus we assumed $3 | (n^3-n) $ is true... But how does step 2 become 3? 


Thanks in advance
 A: $P(n+1)$ is the statement that $(n+1)^3 - (n+1)$ is divisible by 3. Just fill in $n+1$ in $P(n)$. So you have to simplify this expression first.
The equality between 2 and 3 is just algebra (you write $2n$ from the second term as $3n - n$, which is true for all $n$).
Why would you do this? Because you want to use your assumption that $P(n)$ is true, which says that $n^3 - n$ is divisible by 3. The second term has no term $n^3 - n$ but it has an $n^3$ and then substracts $n$ to get the number $n^3 - n$, about which we know something (!), and add $n$ to another term to compensate and keep the equality. 
Now note that the last line is a sum of multiples of 3, the first one, $n^3 - n$, by the inductive assumption.
A: *

*$(n+1)^3-(n+1)$ comes from the expression $n^3-n$ when you replace $n$ with $n+1$. This is the induction step - to show that $(n+1)^3-(n+1)$ is divisable by $3$ if we assume that $n^3-n$ is.

*Here the following happens: $n^3-n$ is divisable by $3$ (by the inductive assumption), $3n^2$ is divisable, $3n$ is divisable, therefore, their sum $n^3-n+3n^2+3n$ is divisable too. (Note that $+2n=-n+3n$ here from 2 to 3.)

