Expansion of $a^x$- $b^x$ I'm sorry if this is too basic but I don't understand why the following is true:
$$
a^3- b^3 = (a - b)(b^2 + ab + a^2)
$$
The professor used this in a proof but I don't see how I would've thought of this. What is the reasoning behind this expansion?
 A: If you multiply out the brackets, you will get
$$
(b-a)(a^2+2ab+b^2) = b^3+2b^2a + ba^2 - b^2a-2ba^2-a^3\\
= b^3 + b^2a-ba^2-a^3
$$
which isn't right.
It's supposed to be
$$
a^3 - b^3 = (a-b)(a^2 +ab + b^2)
$$
which you may (should) verify yourself. This works in general:
$$
a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})
$$
and may (should) be verified through just expanding and cancelling.
A: \begin{eqnarray} a^3-b^3 &=& a^3\color{red}{-a^2b+a^2b}-b^3\\
 &=& (a^3-a^2b)+(a^2b-b^3)\\
&=&...\end{eqnarray}
A: This is a standard factorization and there is a general formula for it:
For $n\in\mathbb{N}$ and $a,b\in\mathbb{R}$ we have
$a^n-b^n=(a-b)\sum_{k=0}^{n-1} a^kb^{n-k-1}$
You can prove this simply by induction. You might try this, it is not that difficult.
Note, that we have for n=2 the known binomial formula:
$a^2-b^2=(a-b)(a+b)$ since $(a-b)\sum_{k=0}^{1} a^kb^{1-k}=(a-b)\cdot (a^0b^1+a^1b^0)=(a-b)(a+b)$
And for $n=3$ we get:
$a^3-b^3=(a-b)(a^2+ab+b^2)$ 
Which formula would we get for $n=4$? You might want to write that out on your own.
Note also, that for the expression $a^n-b^n$ you have a root for $a=b$ so you can do a long division:
$(a^n-b^n)\div (a-b)=\dotso$
And conclute the same result.
There are some nice tricks you can use here. Sometimes the known geometric sum is proven this way:

$\sum_{k=0}^n q^k=\frac{1-q^{n+1}}{1-q}$

If you set the sum
$S=1+q+q^2+\dotso q^n$
And then multiply by $q$ you get:
$qS=q+q^2+\dotso +q^n+q^{n+1}$
If you subtract both equations, we get:
$$S-qS=(1+q+q^2+\dotso +q^n)-(q+q^2+\dotso +q^n+q^{n+1})$$
You see, that most sumands cancel each other out (we say the sum is 'telescoping'). We get:
$$(1-q)S=1-q^{n+1}$$ and so $$S=\frac{1-q^{n+1}}{1-q}$$
The same trick is used when factorizing $a^n-b^n$. 
Note also that $1-q^{n}=1^n-q^n$
