For $-1<r<1$, let $S(r)$ denote the sum of the geometric series \[12+12r+12r^2+12r^3+\cdots .\]  Let $a$ between $-1$ and $1$ satisfy $S(a)S(-a)=2016$. Find $S(a)+S(-a)$.
The sum of an infinite geometric series is $\frac{a}{1-r}\rightarrow \frac{12}{1\mp a}$. The product $S(a)S(-a)=\frac{144}{1-a^2}=2016$. $\frac{12}{1-a}+\frac{12}{1+a}=\frac{24}{1-a^2}$, so the answer is $\frac{2016}{6}=\boxed{336}$.