Binomial Theorem problem!

Hey,

Doing homework, and came across this question. I cannot for the life of me figure out what I am doing wrong, but it doesn't match up to the answer at the back of the book. It does on wolfram alpha. So am I doing this correctly?

**Question:**

Find the first five terms in the expansion of each of the following:

b) $\displaystyle (2a + 3a^{-2})^8 $

**My attempted answer:** ***I do not know how to show combinations using Latex, so hopefully this makes sense. Example: $\displaystyle (8C0) $ is "8 choose 0"***

**First Term:**

$\displaystyle [(8C0)(2a)^8(3a^{-2})^0] $

$\displaystyle = [(1)(256a^8)(1)] $

$\displaystyle = 256a^8 $

**Second Term**:

$\displaystyle [(8C1)(2a)^7(3a^{-2})^1] $

$\displaystyle = [(8)(128a^7)(3a^{-2})] $

$\displaystyle = 3072a^5 $

**Third Term:**

$\displaystyle [(8C2)(2a)^6(3a^{-2})^2] $

$\displaystyle = [(28)(64a^6)(9a^{-4})] $

$\displaystyle = 16128a^2 $

**Fourth Term:**

$\displaystyle [(8C3)(2a)^5(3a^{-2})^3) $

$\displaystyle = [(56)(32a^5)(27a^{-6}) $

$\displaystyle = 48384a^{-1} $

**Fifth Term:**

$\displaystyle [(8C4)(2a)^4(3a^{-2})^4)] $

$\displaystyle [(70)(16a^4)(81a^{-8})] $

$\displaystyle 90720a^{-4} $

Answer at back of book:

$\displaystyle 256a^{-8} + 3072a^{-9} + 16128a^{-10} + 48384a^{-11} + 90720a^{-12} +... $

Wolfram alpha: http://www.wolframalpha.com/input/?i=%282a+%2B+%283a^%28-2%29%29%29^8