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How to simplify $\large{\frac{-y^2}{(y^2-yx)^2}}$?

Here's what I have done:

$= \large{\frac{-y^2}{(y^2-yx)(y^2-yx)}}$

$= \large{\frac{-y^2}{(y-x)y(y-x)y}}$

$= \large{\frac{-y^2}{y^2(y-x)(y-x)}}$

$= \large{\frac{-1}{(y-x)(y-x)}}$

$= \large{\frac{-1}{(y-x)^2}}$

Have I simplified it correctly?

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  • $\begingroup$ Yes, the simplification is correct $\endgroup$
    – user203327
    May 8, 2015 at 16:29

2 Answers 2

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Yes, you did correctly. Another way is $$\frac{-y^2}{(y^2-yx)^2}=-\left(\frac{y}{y^2-yx}\right)^2=-\left(\frac{1}{y-x}\right)^2=-\frac{1}{(y-x)^2}.$$

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Or you could do the following: $$-\frac{y^2}{(y^2 - yx)^2} = -\frac{y^2}{\big(y(y-x)\big)^2} = -\frac{y^2}{y^2(y-x)^2} = -\frac{1}{(y-x)^2},\ y\neq 0, y\neq x$$

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