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:
$$\frac{(x-y)^2-z^2}{(x+y)^2-z^2}\div\frac{x-y+z}{x+y-z}$$
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.
$$\frac{x^2-2xy+y^2-z^2}{x^2+2xy+y^2-z^2}\div\frac{x-y+z}{x+y-z}$$
Then, multiplying the dividend by the reciprocal of the divisor and rewriting.
$$\frac{x^2-2xy+y^2-z^2}{x^2+2xy+y^2-z^2}\times\frac{x+y-z}{x-y+z}$$
$$({x^2-2xy+y^2-z^2})({x+y-z})\over({x^2+2xy+y^2-z^2})({x-y+z})$$
Expanding the expression and grouping i got
$$x^3 - x^2(y + z) - x(y - z)^2 + y^3 - y^2z - yz^2 + z^3 \over x^3 + x^2(y + z) - x(y - z)^2 - y^3 + y^2z + yz^2 - z^3$$
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.
Yuck. You seem to have missed a trick early on. Notice that $x^2 - y^2 = (x+y)(x-y)$. That should help out a lot - before you do anything like expand brackets.