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)

tuzkutimonq4 2022-09-10 Answered
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
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.
x 2 2 x y + y 2 z 2 x 2 + 2 x y + y 2 z 2 ÷ x y + z x + y z
Then, multiplying the dividend by the reciprocal of the divisor and rewriting.
x 2 2 x y + y 2 z 2 x 2 + 2 x y + y 2 z 2 × x + y z x y + z
( x 2 2 x y + y 2 z 2 ) ( x + y z ) ( x 2 + 2 x y + 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 2 z y z 2 + z 3 x 3 + x 2 ( y + z ) x ( y z ) 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:
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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Answers (2)

Willie Gilmore
Answered 2022-09-11 Author has 8 answers
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.
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iescabroussexg
Answered 2022-09-12 Author has 4 answers
Expanding the first term gives
x 2 2 x y + y 2 z 2 x 2 + 2 x z y 2 + z 2 .
Now it easy to see that this is equal to
x y z x + y + z .
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