Simplify $\frac{x^3-x}{x^2+xy+x+y}$ $$\frac{x^3-x}{x^2+xy+x+y}$$
What I did:
 $$\frac{x}{xy+x+y}$$
through simplifying the $x$'s.
But it's not right.  
What did I do wrong?

A good way of checking your working is to put some numbers in. Let's take $x=4,y=2$. Then what you have said is that $$\frac{64-4}{16+4\times2+4+2}=\frac{4}{4\times 2+4+2}$$
but we can see that the first fraction is: \begin{align}\frac{64-4}{16+4\times2+4+2}&=\frac{60}{30}\\
&= 2\end{align} and the second fraction is \begin{align}\frac{4}{4\times2+4+2}&=\frac{4}{14}\\
&= \frac 27\end{align} and so we can very quickly notice that we have done something wrong. 
You will probably have been told that we can't cancel out when there is a $+$, so we can't say that $$\frac {x+y}x = y$$ Well this is the same principle, we just have more $+$'s in the denominator. 
You may remember that, when we have $+$'s in the numerator or the denominator, that we have to cancel the same thing from all terms, so we could say $$\frac{x+xy}{x} = \frac{\color{red}x+\color{red}xy}{\color{red}x} = \frac{1+y}1 = 1+y$$ because we have cancelled an $x$ out of every term in the fraction.
We can apply the same principle here. First, we must factorise the top and the bottom of the fraction, as we cannot cancel the same thing from every term yet.
The top becomes \begin{align}x^3-x &= x(x^2-1)\\&= x(x-1)(x+1)\end{align} (we do the second step by remembering the difference of 2 squares formula) and the bottom becomes \begin{align}x^2+xy+x+y &= (x^2+xy)+(x+y)\\
&= (x(x+y)) + 1(x+y)\\
&= (1+x)(x+y) \end{align}
Therefore, we have $$\frac{x^3-x}{x^2+xy+x+y} = \frac{x(x-1)(x+1)}{(x+1)(x+y)}$$
Now we can say \begin{align}\frac{x(x-1)(x+1)}{(x+1)(x+y)} &= \frac{x(x-1)\color{red}{(x+1)}}{\color{red}{(x+1)}(x+y)}\\
&= \frac{x(x-1)}{(x+y)}\end{align}
