Factorize $ (kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz) $ [closed]

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I tried opening the brackets but it keeps getting complex and complex. Does anyone have any easier way of doing it. Please tell how did you observe.







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closed as off-topic by Alex Francisco, Xander Henderson, Leucippus, Shailesh, Parcly Taxel Jul 17 at 3:53


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Xander Henderson, Leucippus, Shailesh, Parcly Taxel
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    up vote
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    favorite












    I tried opening the brackets but it keeps getting complex and complex. Does anyone have any easier way of doing it. Please tell how did you observe.







    share|cite|improve this question











    closed as off-topic by Alex Francisco, Xander Henderson, Leucippus, Shailesh, Parcly Taxel Jul 17 at 3:53


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Xander Henderson, Leucippus, Shailesh, Parcly Taxel
    If this question can be reworded to fit the rules in the help center, please edit the question.














      up vote
      -1
      down vote

      favorite









      up vote
      -1
      down vote

      favorite











      I tried opening the brackets but it keeps getting complex and complex. Does anyone have any easier way of doing it. Please tell how did you observe.







      share|cite|improve this question











      I tried opening the brackets but it keeps getting complex and complex. Does anyone have any easier way of doing it. Please tell how did you observe.









      share|cite|improve this question










      share|cite|improve this question




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      asked Jul 16 at 17:10









      Harsh Katara

      458




      458




      closed as off-topic by Alex Francisco, Xander Henderson, Leucippus, Shailesh, Parcly Taxel Jul 17 at 3:53


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Xander Henderson, Leucippus, Shailesh, Parcly Taxel
      If this question can be reworded to fit the rules in the help center, please edit the question.




      closed as off-topic by Alex Francisco, Xander Henderson, Leucippus, Shailesh, Parcly Taxel Jul 17 at 3:53


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Xander Henderson, Leucippus, Shailesh, Parcly Taxel
      If this question can be reworded to fit the rules in the help center, please edit the question.




















          2 Answers
          2






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          $$ f(x,y,z,k)=(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz) tag 1$$
          Name the six terms respectively as :
          $$f(x,y,z,k)=A*B*C-a*b*c$$
          If $k=1$ we have $A=c$ , $B=b$ and $C=b$ thus $ABC=abc$ and $f(x,y,z,1)=0$. As a consequence $(k-1)$ is a factor of $f(x,y,z,k)$.



          Do the same with $k=-1$ which leads to $f(x,y,z,-1)=0$ then a second factor $(k+1)$.



          Proceeding on the same manner successively with $x=y$ , $y=z$ , and $z=x$ gives three other factors : $(x-y)$ , $(y-z)$ and $(z-x)$.



          Finally
          $$f(x,y,z,k)=-2(k-1)(k+1)(x-y)(y-z)(z-x)tag 2$$



          The coefficient $-2$ is obtained in giving an arbitrary value to each $x,y,z,k$ , of course so that $fneq 0$ , and comparing the values of Eq.$(1)$ and Eq.$(2)$.






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            Hint
            Let $$P(x,y,z,k)=(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz).$$
            Prove that $$P(x,x,z,k)=P(x,y,x,k)=P(x,y,y,k)=P(x,y,z,pm 1)=0.$$






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            • This is a nice hint. Do you have any advice on how one would recognize that this is the right way to approach the problem? Are there more computational ways to do it?
              – user219923
              Jul 16 at 17:58

















            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            $$ f(x,y,z,k)=(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz) tag 1$$
            Name the six terms respectively as :
            $$f(x,y,z,k)=A*B*C-a*b*c$$
            If $k=1$ we have $A=c$ , $B=b$ and $C=b$ thus $ABC=abc$ and $f(x,y,z,1)=0$. As a consequence $(k-1)$ is a factor of $f(x,y,z,k)$.



            Do the same with $k=-1$ which leads to $f(x,y,z,-1)=0$ then a second factor $(k+1)$.



            Proceeding on the same manner successively with $x=y$ , $y=z$ , and $z=x$ gives three other factors : $(x-y)$ , $(y-z)$ and $(z-x)$.



            Finally
            $$f(x,y,z,k)=-2(k-1)(k+1)(x-y)(y-z)(z-x)tag 2$$



            The coefficient $-2$ is obtained in giving an arbitrary value to each $x,y,z,k$ , of course so that $fneq 0$ , and comparing the values of Eq.$(1)$ and Eq.$(2)$.






            share|cite|improve this answer

























              up vote
              2
              down vote



              accepted










              $$ f(x,y,z,k)=(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz) tag 1$$
              Name the six terms respectively as :
              $$f(x,y,z,k)=A*B*C-a*b*c$$
              If $k=1$ we have $A=c$ , $B=b$ and $C=b$ thus $ABC=abc$ and $f(x,y,z,1)=0$. As a consequence $(k-1)$ is a factor of $f(x,y,z,k)$.



              Do the same with $k=-1$ which leads to $f(x,y,z,-1)=0$ then a second factor $(k+1)$.



              Proceeding on the same manner successively with $x=y$ , $y=z$ , and $z=x$ gives three other factors : $(x-y)$ , $(y-z)$ and $(z-x)$.



              Finally
              $$f(x,y,z,k)=-2(k-1)(k+1)(x-y)(y-z)(z-x)tag 2$$



              The coefficient $-2$ is obtained in giving an arbitrary value to each $x,y,z,k$ , of course so that $fneq 0$ , and comparing the values of Eq.$(1)$ and Eq.$(2)$.






              share|cite|improve this answer























                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                $$ f(x,y,z,k)=(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz) tag 1$$
                Name the six terms respectively as :
                $$f(x,y,z,k)=A*B*C-a*b*c$$
                If $k=1$ we have $A=c$ , $B=b$ and $C=b$ thus $ABC=abc$ and $f(x,y,z,1)=0$. As a consequence $(k-1)$ is a factor of $f(x,y,z,k)$.



                Do the same with $k=-1$ which leads to $f(x,y,z,-1)=0$ then a second factor $(k+1)$.



                Proceeding on the same manner successively with $x=y$ , $y=z$ , and $z=x$ gives three other factors : $(x-y)$ , $(y-z)$ and $(z-x)$.



                Finally
                $$f(x,y,z,k)=-2(k-1)(k+1)(x-y)(y-z)(z-x)tag 2$$



                The coefficient $-2$ is obtained in giving an arbitrary value to each $x,y,z,k$ , of course so that $fneq 0$ , and comparing the values of Eq.$(1)$ and Eq.$(2)$.






                share|cite|improve this answer













                $$ f(x,y,z,k)=(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz) tag 1$$
                Name the six terms respectively as :
                $$f(x,y,z,k)=A*B*C-a*b*c$$
                If $k=1$ we have $A=c$ , $B=b$ and $C=b$ thus $ABC=abc$ and $f(x,y,z,1)=0$. As a consequence $(k-1)$ is a factor of $f(x,y,z,k)$.



                Do the same with $k=-1$ which leads to $f(x,y,z,-1)=0$ then a second factor $(k+1)$.



                Proceeding on the same manner successively with $x=y$ , $y=z$ , and $z=x$ gives three other factors : $(x-y)$ , $(y-z)$ and $(z-x)$.



                Finally
                $$f(x,y,z,k)=-2(k-1)(k+1)(x-y)(y-z)(z-x)tag 2$$



                The coefficient $-2$ is obtained in giving an arbitrary value to each $x,y,z,k$ , of course so that $fneq 0$ , and comparing the values of Eq.$(1)$ and Eq.$(2)$.







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 16 at 18:47









                JJacquelin

                40.1k21649




                40.1k21649




















                    up vote
                    0
                    down vote













                    Hint
                    Let $$P(x,y,z,k)=(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz).$$
                    Prove that $$P(x,x,z,k)=P(x,y,x,k)=P(x,y,y,k)=P(x,y,z,pm 1)=0.$$






                    share|cite|improve this answer























                    • This is a nice hint. Do you have any advice on how one would recognize that this is the right way to approach the problem? Are there more computational ways to do it?
                      – user219923
                      Jul 16 at 17:58














                    up vote
                    0
                    down vote













                    Hint
                    Let $$P(x,y,z,k)=(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz).$$
                    Prove that $$P(x,x,z,k)=P(x,y,x,k)=P(x,y,y,k)=P(x,y,z,pm 1)=0.$$






                    share|cite|improve this answer























                    • This is a nice hint. Do you have any advice on how one would recognize that this is the right way to approach the problem? Are there more computational ways to do it?
                      – user219923
                      Jul 16 at 17:58












                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    Hint
                    Let $$P(x,y,z,k)=(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz).$$
                    Prove that $$P(x,x,z,k)=P(x,y,x,k)=P(x,y,y,k)=P(x,y,z,pm 1)=0.$$






                    share|cite|improve this answer















                    Hint
                    Let $$P(x,y,z,k)=(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz).$$
                    Prove that $$P(x,x,z,k)=P(x,y,x,k)=P(x,y,y,k)=P(x,y,z,pm 1)=0.$$







                    share|cite|improve this answer















                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jul 16 at 17:51


























                    answered Jul 16 at 17:27









                    Leox

                    5,0921323




                    5,0921323











                    • This is a nice hint. Do you have any advice on how one would recognize that this is the right way to approach the problem? Are there more computational ways to do it?
                      – user219923
                      Jul 16 at 17:58
















                    • This is a nice hint. Do you have any advice on how one would recognize that this is the right way to approach the problem? Are there more computational ways to do it?
                      – user219923
                      Jul 16 at 17:58















                    This is a nice hint. Do you have any advice on how one would recognize that this is the right way to approach the problem? Are there more computational ways to do it?
                    – user219923
                    Jul 16 at 17:58




                    This is a nice hint. Do you have any advice on how one would recognize that this is the right way to approach the problem? Are there more computational ways to do it?
                    – user219923
                    Jul 16 at 17:58


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