Why is $|x^2 - a^2| = |x - a||x + a|$? By the definition of absolute value, we have
\begin{equation}
|x^2 - a^2| =
\begin{cases}
(x - a)(x + a) \ if \ x^2 \geq a^2\\
-(x - a)(x + a) \ if \ x^2 \leq a^2.
\end{cases}
\end{equation}
How do we conclude that the right hand side equals $|x - a||x + a|$?

Assume that $a\ge0$.
Note that $x^2\ge a^2$ if and only if $x\le -a$ or $x\ge a$.
If $x\le-a$, then $x+a\le0$ and $x-a\le0$. So $|x+a||x-a|=[-(x+a)][-(x-a)]=(x+a)(x-a)$.
If $x\ge a$, then $x+a\ge0$ and $x-a\ge0$. So $|x+a||x-a|=(x+a)(x-a)$.
We also have $x^2\le a^2$ if and only if $-a\le x\le a$. In such case, $x+a\ge0$ and $x-a\le 0$. So $|x+a||x-a|=(x+a)[-(x-a)]=-(x+a)(x-a)$.
Therefore, $|x^2-a^2|=|x+a||x-a|$.
If $a<0$, then $-a>0$. So we have $|x^2-a^2|=|x^2-(-a)^2|=|x+(-a)||x-(-a)|=|x-a||x+a|$.
