# Advice on simple exercise; dividing rational expression I'm looking into rational expressions to clarify some doubts that i have. I'm solving the exercises at the end of the chapter and I'm stuck with one of them. This is it: ((x-y)^2-z^2)/((x+y)^2-z^2)-:(x-y+z)/(x+y-z)

Advice on simple exercise; dividing rational expression
I'm looking into rational expressions to clarify some doubts that i have. I'm solving the exercises at the end of the chapter and I'm stuck with one of them. This is it:
$\frac{\left(x-y{\right)}^{2}-{z}^{2}}{\left(x+y{\right)}^{2}-{z}^{2}}÷\frac{x-y+z}{x+y-z}$
I have been solving several of this type by:
1.- Factoring out the largest common factor if possible.
2.- Factoring.
3.- Removing factors of 1.
In this case, i'm trying to find a way of solving this without expanding the hole thing, but i have not found the way jet.
I start by expanding both square binomials of the dividend.
$\frac{{x}^{2}-2xy+{y}^{2}-{z}^{2}}{{x}^{2}+2xy+{y}^{2}-{z}^{2}}÷\frac{x-y+z}{x+y-z}$
Then, multiplying the dividend by the reciprocal of the divisor and rewriting.
$\frac{{x}^{2}-2xy+{y}^{2}-{z}^{2}}{{x}^{2}+2xy+{y}^{2}-{z}^{2}}×\frac{x+y-z}{x-y+z}$
$\frac{\left({x}^{2}-2xy+{y}^{2}-{z}^{2}\right)\left(x+y-z\right)}{\left({x}^{2}+2xy+{y}^{2}-{z}^{2}\right)\left(x-y+z\right)}$
Expanding the expression and grouping i got
$\frac{{x}^{3}-{x}^{2}\left(y+z\right)-x\left(y-z{\right)}^{2}+{y}^{3}-{y}^{2}z-y{z}^{2}+{z}^{3}}{{x}^{3}+{x}^{2}\left(y+z\right)-x\left(y-z{\right)}^{2}-{y}^{3}+{y}^{2}z+y{z}^{2}-{z}^{3}}$
The problem is that i don't know what to do now. I know that the result is:
$\frac{x-y-z}{x+y+z}$
I want to belive that there's an easy way from the beggining (to avoid expanding all the thing) but i would like your help.
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Willie Gilmore
Yuck. You seem to have missed a trick early on. Notice that ${x}^{2}-{y}^{2}=\left(x+y\right)\left(x-y\right)$. That should help out a lot - before you do anything like expand brackets.
###### Did you like this example?
iescabroussexg
Expanding the first term gives
$\frac{{x}^{2}-2xy+{y}^{2}-{z}^{2}}{{x}^{2}+2xz-{y}^{2}+{z}^{2}}.$
Now it easy to see that this is equal to
$\frac{x-y-z}{x+y+z}.$