Factoring $a^3-b^3$ I recently got interested in mathematics after having slacked through it in highschool. Therefore I picked up the book "Algebra" by I.M. Gelfand and A.Shen
At problem 113, the reader is asked to factor $a^3-b^3.$
The given solution is: 
$$a^3-b^3 = a^3 - a^2b + a^2b -ab^2 + ab^2 -b^3 = a^2(a-b) + ab(a-b) + b^2(a-b) = (a-b)(a^2+ab+b^2)$$
I was wondering how the second equality is derived. From what is it derived, from $a^2-b^2$? I know that the result is the difference of cubes formula, however searching for it on the internet i only get exercises where the formula is already given. Can someone please point me in the right direction?

The second equality is obtained by first grouping terms in the middle expression and then factoring the grouped terms:
$$\begin{align}
a^3-a^2b+a^2b-ab^2+ab^2-b^3&=(a^3-a^2b)+(a^2b-ab^2)+(ab^2-b^3)\cr
&=a^2(a-b)+ab(a-b)+b^2(a-b)\cr
\end{align}$$
If the OP is wondering where the middle expression came from in the first place, it's kind of a deus ex machina:  All you're doing is sticking two $0$'s between $a^3$ and $-b^3$ (the expressions $-a^2b+a^2b$ and $-ab^2+ab^2$ are both obviously equal to $0$), but when you do, lo and behold, the regrouping and factoring work their magic.  
