Exponent questions algebra How would I solve the following two exponent questions?
(1) The first question is 
$$\left(\frac{x^{-2}+y^{-1}}{xy^2}\right)^{-1}$$
I got $\quad \displaystyle \frac{-xy^{-2}}{x^2+y},\;\;$but this does not seem to be correct.
(2) My second question is 
$$\left(\frac{3}{A^{-3}B^{-2}}\right)^{-2}$$
I got $\quad\displaystyle \frac{A^6B^4}{1/9},\quad$ but my book's answer is $\quad\displaystyle \frac{1}{9A^6B^4}$
 A: We are not solving, we are simplifying: For the first, note that 

$$\left(\frac{x^{-2}+y^{-1}}{xy^2}\right)^{-1}\;=\; \frac{xy^2}{x^{-2} + y^{-1}}\;=\;\frac{xy^2}{\large\frac{1}{x^2} + \frac{1}{y}}$$

Try now to multiply numerator and denominator by $x^2y$:
$$\frac{xy^2}{\left(\large\frac{1}{x^2} + \frac{1}{y}\right)}\cdot \frac{(x^2y)}{(x^2y)} \quad = \quad\frac {x\cdot x^2 \cdot y^2 \cdot y}{\left(\large\frac{x^2y}{x^2} + \frac{x^2y}{y}\right)}\quad =\quad \frac{x^3y^3}{y + x^2} $$

For the second, again, we are simplifying:
$$\left(\frac{3}{A^{-3}B^{-2}}\right)^{-2} \quad = \quad \frac{3^{-2}}{A^{(-3)(-2)}B^{(-2)(-2)}}\quad = \quad \frac{1/9}{A^6B^4}\quad =\quad \frac{1}{9A^6B^4}$$

(as per my now deleted comment below) Alternatively, for the second problem, we proceed as follows:
$$\left(\frac{3}{A^{-3}B^{-2}}\right)^{-2}\quad = \quad\left(\frac{A^{-3}B^{-2}}{3}\right)^2 \quad=\quad \frac{A^{-6}B^{-4}}{9} \quad = \quad \frac{1}{9A^6B^4}$$
A: You got 

$$\frac{-xy^{-2}}{x^2+y}$$

Here's what I got:
$$\left(\frac{x^{-2}+y^{-1}}{xy^2}\right)^{-1}=\left(\frac{xy^2}{x^{-2}+y^{-1}}\right)=\left(\frac{xy^2}{\frac{1}{x^2}+\frac{1}{y}} \right)=\left(\frac{xy^2}{\frac{y+x^2}{yx^2}} \right)=xy^2\cdot\left(\frac{yx^2}{y+x^2} \right)=\frac{x^3y^3}{y+x^2}$$
A: To Solve Problem 1. 
We are going to begin by seperating the fraction into two parts (both still underneath one big -1st power)
This leads us to x^-3/y^2 + y^-3/x. After moving the powers to their correct locations we get:
1/(y^2*x^3) + 1(x*y^3).
Now factoring out 1/(x*y^2) from both sides we get.
1/(x*y^2)*[1/(x^2) + 1/(y)]
finding an LCD (least common denominator) for both fractions we can rewrite it as.
(1/x*y^2)*[(y + x^2)/(x^2*y)]
Placing back the factored fraction we end up with:
(y + x^2)/(x^3 * y^3)
and now applying the negative first power (REMEMBER FROM THE START!)
(x^3*y^3)/(y + x^2)
Is the correct answer.
To Solve Number 2.
We begin with the expression (3/(A^(-3) * B^(-2)))^(-2).
First we bring up the A and B (since we can w/ negative exponents!)
so we get (3*A^3 * B^2)^(-2). Now we can evaluate that -2nd power in two steps. First we evaluate it as a 2nd power and then a -1, both we know how to do.
So after evaluating the 2 we end up with:
(9*A^6*B^4)^(-1)
After the -1 we end up with:
1/(9*A^6*B^4)
Which is what you should get! Good Luck :)
A: Here is another way to simplify the second expression from your question:
$$\left(\frac{3}{A^{-3}B^{-2}}\right)^{-2} = \left(\frac{A^{-3}B^{-2}}{3}\right)^{2} = \left(\frac{A^{-6}B^{-4}}{9}\right) = \left(\frac{1}{9A^{6}B^{4}}\right).$$
