Compute $frac{x}{y-z} +frac {y}{z-x}+frac {z}{x-y} $












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Compute $frac{x}{y-z} +frac {y}{z-x}+frac {z}{x-y} $ if $frac{x^2}{ (y-z)^2} +frac {y^2}{ (z-x)^2}+frac {z^2}{ (x-y)^2}=11. $



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algebra-precalculus summation factoring







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    $begingroup$


    Compute $frac{x}{y-z} +frac {y}{z-x}+frac {z}{x-y} $ if $frac{x^2}{ (y-z)^2} +frac {y^2}{ (z-x)^2}+frac {z^2}{ (x-y)^2}=11. $



    I have no idea how can I start.






    algebra-precalculus summation factoring







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      $begingroup$


      Compute $frac{x}{y-z} +frac {y}{z-x}+frac {z}{x-y} $ if $frac{x^2}{ (y-z)^2} +frac {y^2}{ (z-x)^2}+frac {z^2}{ (x-y)^2}=11. $



      I have no idea how can I start.






      algebra-precalculus summation factoring







      share|cite|improve this question











      $endgroup$




      Compute $frac{x}{y-z} +frac {y}{z-x}+frac {z}{x-y} $ if $frac{x^2}{ (y-z)^2} +frac {y^2}{ (z-x)^2}+frac {z^2}{ (x-y)^2}=11. $



      I have no idea how can I start.






      algebra-precalculus summation factoring




      algebra-precalculus summation factoring






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      edited Jan 6 at 20:21









      Michael Rozenberg

      99.3k1590189




      99.3k1590189










      asked Jan 6 at 17:39









      rafarafa

      598212




      598212






















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          $$left(sum_{cyc}frac{x}{y-z}right)^2=sum_{cyc}frac{x^2}{(y-z)^2}+2sum_{cyc}frac{xy}{(y-z)(z-x)}=$$
          $$=11+2sum_{cyc}frac{xy(x-y)}{(x-y)(y-z)(z-x)}=11+2frac{sumlimits_{cyc}(x^2y-x^2z)}{(x-y)(y-z)(z-x)}=$$
          $$=11-frac{2(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x)}=9.$$






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            $begingroup$

            $$left(sum_{cyc}frac{x}{y-z}right)^2=sum_{cyc}frac{x^2}{(y-z)^2}+2sum_{cyc}frac{xy}{(y-z)(z-x)}=$$
            $$=11+2sum_{cyc}frac{xy(x-y)}{(x-y)(y-z)(z-x)}=11+2frac{sumlimits_{cyc}(x^2y-x^2z)}{(x-y)(y-z)(z-x)}=$$
            $$=11-frac{2(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x)}=9.$$






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              5












              $begingroup$

              $$left(sum_{cyc}frac{x}{y-z}right)^2=sum_{cyc}frac{x^2}{(y-z)^2}+2sum_{cyc}frac{xy}{(y-z)(z-x)}=$$
              $$=11+2sum_{cyc}frac{xy(x-y)}{(x-y)(y-z)(z-x)}=11+2frac{sumlimits_{cyc}(x^2y-x^2z)}{(x-y)(y-z)(z-x)}=$$
              $$=11-frac{2(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x)}=9.$$






              share|cite|improve this answer









              $endgroup$
















                5












                5








                5





                $begingroup$

                $$left(sum_{cyc}frac{x}{y-z}right)^2=sum_{cyc}frac{x^2}{(y-z)^2}+2sum_{cyc}frac{xy}{(y-z)(z-x)}=$$
                $$=11+2sum_{cyc}frac{xy(x-y)}{(x-y)(y-z)(z-x)}=11+2frac{sumlimits_{cyc}(x^2y-x^2z)}{(x-y)(y-z)(z-x)}=$$
                $$=11-frac{2(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x)}=9.$$






                share|cite|improve this answer









                $endgroup$



                $$left(sum_{cyc}frac{x}{y-z}right)^2=sum_{cyc}frac{x^2}{(y-z)^2}+2sum_{cyc}frac{xy}{(y-z)(z-x)}=$$
                $$=11+2sum_{cyc}frac{xy(x-y)}{(x-y)(y-z)(z-x)}=11+2frac{sumlimits_{cyc}(x^2y-x^2z)}{(x-y)(y-z)(z-x)}=$$
                $$=11-frac{2(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x)}=9.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 6 at 19:33









                Michael RozenbergMichael Rozenberg

                99.3k1590189




                99.3k1590189






























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