The fraction
\[\frac{\left(3^{2008}\right)^2-\left(3^{2006}\right)^2}{\left(3^{2007}\right)^2-\left(3^{2005}\right)^2}\]
simplifies to which of the following?
$\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ \frac{9}{4}\qquad\mathrm{(C)}\ 3\qquad\mathrm{(D)}\ \frac{9}{2}\qquad\mathrm{(E)}\ 9$
Simplifying, we get \[\frac{3^{4016}-3^{4012}}{3^{4014}-3^{4010}}.\] Factoring out $3^{4012}$ in the numerator and factoring out $3^{4010}$ in the denominator gives us \[\frac{(3^4-1)(3^{4012})}{(3^4-1)(3^{4010})}.\] Canceling out $3^4-1$ gives us $\frac{3^{4012}}{3^{4010}}=\frac{3^2}{3^0}=9\ \mathrm{(E)}.$