<s>[INST] 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 $\quad\displaystyle \frac{1}{9A^6B^4}$
 [/INST] 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:
$$\left(\frac{3}{A^{-3}B^{-2}}\right)^{-2}\quad = \quad\left(\frac{A^{-3}B^{-2}}{3}\right)^2 \quad=\quad \frac{A^{-6}B^{-4}}{9} \quad = \quad \frac{1}{9A^6B^4}$$
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