Factoring and expanding $1-x^8$ It's been a while since I've studied factoring, and I need it for a question. How did this go from $1-x^8$ to $(1-x^4)(1+x^4)$ and then to $(1-x)(1+x+...+x^7)$? I remember studying this a few years go but unfortunately I don't remember.
$$1-x^8=(1-x^4)(1+x^4)=(1-x)(1+x+\dots + x^7)$$
 A: You may remember $$(a+b)(a-b) = a^2 - b^2.$$  To see this, just multiply out.  You get $a^2 + ba - ab - b^2$, and the middle terms cancel, leaving you with $a^2-b^2$.
Then if you take $a=1, b=x^4$, you get $$(1+x^4)(1-x^4) = 1-x^8.$$  That's your first identity.
You may also remember $$(1+x+x^2+\ldots+x^n)(x-1) = x^{n+1}-1.$$  It's possible that you remember the formula for the sum of a geometric series:  $$1+x+x^2+\ldots+x^n = \frac{x^{n+1}-1}{x-1}$$
but if not, just do the multiplication: distribute the right-hand factor $x-1$ over the left-hand factor:
$$\begin{align}
(1+x+x^2+\ldots+x^n)(x-1) 
& = (x-1) + x(x-1) + x^2(x-1) + \ldots + x^n(x-1) \\
& = (\color{darkgreen}{x} - 1)  + (\color{maroon}{x^2}-\color{darkgreen}{x}) + (\color{darkblue}{x^3}-\color{maroon}{x^2}) + \ldots + (x^{n+1}-x^n)  \tag{$\star$}\\
& = x^{n+1} -1
\end{align}$$
Notice that in line $(\star)$ the two $\color{darkgreen}{x}$ terms cancel, the two $\color{maroon}{x^2}$ terms cancel, and so on, leaving only $-1$ and $x^{n+1}$.  Taking $n=7$ we get $$(1+x+x^2+\ldots+x^7)(x-1) = x^8-1.$$
Does this help?
A: Taking your question, “How did this go from $1-x^8$ to $(1-x^4)(1+x^4)$ and then to...” literally, it involves repeated applications of the identity $1-z^2=(1-z)(1+z)$: $$\begin{align} 1+x^8 &= (1-x^4)(1+x^4) \\ &= (1-x^2)(1+x^2)(1+x^4) \\ &= (1-x)(1+x)(1+x^2)(1+x^4) \\ &= (1-x)(1+x^2+x^3+x^4+x^5+x^6+x^7). \end{align}$$
