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$$\frac{x^{-2}-y^{-2}}{x^{-1}+y^{-1}}$$

I know that a fractional expression with negative exponents can be flipped to get the positive exponent but I am not sure that I am taking the correct approach to solving this problem. Every solution that I get doesn't match the textbook's correct answer.

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    $\begingroup$ $\dfrac{a^2-b^2}{a+b} = a-b$, so if $a=x^{-1}, b=y^{-1}$, then expression is equal to $x^{-1}-y^{-1}$. $\endgroup$ – Oleg567 Aug 18 '14 at 11:48
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    $\begingroup$ One possible way: The numerator is the difference of 2 squares. Factorize to get $(x^{-1}-y^{-1})(x^{-1}+y^{-1})$. So, $\frac{x^{-2}-y^{-2}}{x^{-1}+y^{-1}}=\frac{(x^{-1}-y^{-1})(x^{-1}+y^{-1})}{x^{-1}+y^{-1}}=x^{-1}-y^{-1}=\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}$. $\endgroup$ – Radz Aug 18 '14 at 11:52
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$$\begin{align} \frac{x^{-2}-y^{-2}}{x^{-1}+y^{-1}} &=\frac{\frac{1}{x^2}-\frac{1}{y^2}}{\frac{1}{x}+\frac{1}{y}}\\[10pt] &=\frac{y^2-x^2}{x^2y^2} \cdot \frac{xy}{y+x}\\[10pt] &=\frac{(y-x)(y+x)}{xy(y+x)}\\[10pt] &=\frac{y-x}{xy} \end{align}$$

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Multiply the top and bottom by $x^2y^2$ and this will make your fraction a whole lot easier to look at.

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