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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Advice on simple exercise; dividing rational expression I&#039;m looking into rational expressions to clarify some doubts that i have. I&#039;m solving the exercises at the end of the chapter and I&#039;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&#039;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&#039;t know what to do now. I know that the result is:            x      −      y      −      z              x      +      y      +      z      I want to belive that there&#039;s an easy way from the beggining (to avoid expanding all the thing) but i would like your help.

Willie Gilmore · Accepted Answer

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.

iescabroussexg · Accepted Answer

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        .

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)

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