In the equation:
(2n^3)^4 (n^-2)^-1
I work it and get 16n^12-n^2. I am not sure what to do with the "-n^2". Help woulld be greatly appreciated.
First, although it may seem "picky", you don't have an equation here, you have an expression. (It isn't equal to anything.)
What do you want to do with the $\displaystyle n^2$? That is, what are you trying to do with the original expression? Simplify it? $\displaystyle 16n^{12}- n^2$ seems simple enough to me! You could, if you wanted, factor out an "$\displaystyle n^2$". $\displaystyle n^{12}= n^{10}n^2$ so [tex]16n^{12}- n^2= n^2(16n^{10}- 1). If you really wanted to you could factor that further- $\displaystyle 16n^10- 1= (4n^5)^2- 1^2$ is the difference of two squares and can be factored as $\displaystyle (4n^5-1)(4n^5+1)$. whether $\displaystyle n^2(4n^5-1)(4n^5+1)$ is "simpler" than $\displaystyle 16n^{12}- n^2$ is a matter of taste or, perhaps better, of what you wanted to do with it from here.