2008 AMC 10A Problem 7

The exponents are huge, so do not try to compute anything. Instead, write every term using one small building block.

1. Let $x = 3^{2005}$, the smallest power in the problem. Then each other power is $x$ times a few extra $3$'s:
$$3^{2006} = 3x,\qquad 3^{2007} = 9x,\qquad 3^{2008} = 27x.$$
2. Rewrite the top of the fraction:
$$(27x)^2 - (3x)^2 = 729x^2 - 9x^2 = 720x^2.$$
3. Rewrite the bottom:
$$(9x)^2 - x^2 = 81x^2 - x^2 = 80x^2.$$
4. Divide. The $x^2$ cancels:
$$\frac{720x^2}{80x^2} = \frac{720}{80} = 9.$$

The answer is $\boxed{\textbf{(E)}\ 9}$.

Top and bottom are built the same way; the top just uses powers of $3$ that are one step higher. Each step up multiplies a term by $3$, and after squaring by $9$, so the whole top is $9$ times the whole bottom. When you see giant exponents that are close to each other, name the smallest power as a letter and write the rest as small multiples of it.

A quick check with small numbers shows the pattern. Replace the exponents $2008, 2006, 2007, 2005$ by $3, 1, 2, 0$ (each lowered by $2005$):
$$\frac{(3^3)^2 - (3^1)^2}{(3^2)^2 - (3^0)^2} = \frac{729 - 9}{81 - 1} = \frac{720}{80} = 9.$$
Lowering all exponents by the same amount divides top and bottom by the same number, so the ratio does not change.

Rewrite each square: $(3^{2008})^2 = 3^{4016}$, $(3^{2006})^2 = 3^{4012}$, $(3^{2007})^2 = 3^{4014}$, $(3^{2005})^2 = 3^{4010}$.

Factor the smallest power out of each difference:
$$\frac{3^{4012}\left(3^{4} - 1\right)}{3^{4010}\left(3^{4} - 1\right)}.$$

The factors $3^4 - 1 = 80$ cancel, leaving $\frac{3^{4012}}{3^{4010}} = 3^2 = 9$.

The answer is $\boxed{\textbf{(E)}\ 9}$.