2008 AMC 10A Problem 7

Since $\left(3^{k}\right)^2 = 9^{k},$ the fraction is $\dfrac{9^{2008} - 9^{2006}}{9^{2007} - 9^{2005}}.$ Factoring $9^{2005}$ from each part gives $$\dfrac{9^{2005}\left(9^3 - 9\right)}{9^{2005}\left(9^2 - 1\right)} = \dfrac{9\left(9^2 - 1\right)}{9^2 - 1} = 9.$$ Thus, the correct answer is E.