$x^3-x+y=0$ and $y^3+x-y=0$ then what will be the value of x and y? [closed]












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If $x^3-x+y=0$ and $y^3+x-y=0$ then what will be the value of x and y?







linear-algebra







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closed as off-topic by Adrian Keister, Dietrich Burde, Davide Giraudo, Shailesh, Will Fisher Dec 20 '18 at 1:06


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


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    If $x^3-x+y=0$ and $y^3+x-y=0$ then what will be the value of x and y?







    linear-algebra







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    closed as off-topic by Adrian Keister, Dietrich Burde, Davide Giraudo, Shailesh, Will Fisher Dec 20 '18 at 1:06


    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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, Dietrich Burde, Davide Giraudo, Shailesh, Will Fisher

    If this question can be reworded to fit the rules in the help center, please edit the question.



















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      If $x^3-x+y=0$ and $y^3+x-y=0$ then what will be the value of x and y?







      linear-algebra







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      If $x^3-x+y=0$ and $y^3+x-y=0$ then what will be the value of x and y?







      linear-algebra




      linear-algebra






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 19 '18 at 19:07









      greedoid

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      45.3k1159113










      asked Dec 19 '18 at 19:05









      Basant ThakurBasant Thakur

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      closed as off-topic by Adrian Keister, Dietrich Burde, Davide Giraudo, Shailesh, Will Fisher Dec 20 '18 at 1:06


      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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, Dietrich Burde, Davide Giraudo, Shailesh, Will Fisher

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by Adrian Keister, Dietrich Burde, Davide Giraudo, Shailesh, Will Fisher Dec 20 '18 at 1:06


      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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, Dietrich Burde, Davide Giraudo, Shailesh, Will Fisher

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          3 Answers
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          $begingroup$

          Adding two equations, we have $x^3+y^3=0to x^3=-y^3 to x=-y$ (I suppose that $x,yinmathbb{R}$).



          Now, replacing,



          $x^3-x+y=x^3-x-x=0to x^3-2x=0to x^3=2xto x=0$ or $x=pmsqrt{2}$ and $y=0$ or $y=mpsqrt{2}$, respectively.






          share|cite|improve this answer









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            0












            $begingroup$

            Hint: With $$x=y-y^3$$ we get $$(y-y^3)^3-y+y^3+y=0$$ or $$(y-y^3)^3+y^3=0$$
            factorizing we obtain
            $$y^3(y^3-2)(y^4-y^2+1)=0$$ Can you proceed now?






            share|cite|improve this answer









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              $begingroup$

              Since $$x^3+y^3=0 implies (x+y)(x^2-xy+y^2)=0 $$



              and $x^2-xy+y^2ne 0$ we have $x=-y$.



              So $x^3-2x=0$ so $x=0$ (and $y=0$) or $x=pm sqrt{2} $ (and $y=mp sqrt{2} $ )






              share|cite|improve this answer









              $endgroup$




















                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                1












                $begingroup$

                Adding two equations, we have $x^3+y^3=0to x^3=-y^3 to x=-y$ (I suppose that $x,yinmathbb{R}$).



                Now, replacing,



                $x^3-x+y=x^3-x-x=0to x^3-2x=0to x^3=2xto x=0$ or $x=pmsqrt{2}$ and $y=0$ or $y=mpsqrt{2}$, respectively.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Adding two equations, we have $x^3+y^3=0to x^3=-y^3 to x=-y$ (I suppose that $x,yinmathbb{R}$).



                  Now, replacing,



                  $x^3-x+y=x^3-x-x=0to x^3-2x=0to x^3=2xto x=0$ or $x=pmsqrt{2}$ and $y=0$ or $y=mpsqrt{2}$, respectively.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Adding two equations, we have $x^3+y^3=0to x^3=-y^3 to x=-y$ (I suppose that $x,yinmathbb{R}$).



                    Now, replacing,



                    $x^3-x+y=x^3-x-x=0to x^3-2x=0to x^3=2xto x=0$ or $x=pmsqrt{2}$ and $y=0$ or $y=mpsqrt{2}$, respectively.






                    share|cite|improve this answer









                    $endgroup$



                    Adding two equations, we have $x^3+y^3=0to x^3=-y^3 to x=-y$ (I suppose that $x,yinmathbb{R}$).



                    Now, replacing,



                    $x^3-x+y=x^3-x-x=0to x^3-2x=0to x^3=2xto x=0$ or $x=pmsqrt{2}$ and $y=0$ or $y=mpsqrt{2}$, respectively.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 19 '18 at 19:12









                    Martín Vacas VignoloMartín Vacas Vignolo

                    3,816623




                    3,816623























                        0












                        $begingroup$

                        Hint: With $$x=y-y^3$$ we get $$(y-y^3)^3-y+y^3+y=0$$ or $$(y-y^3)^3+y^3=0$$
                        factorizing we obtain
                        $$y^3(y^3-2)(y^4-y^2+1)=0$$ Can you proceed now?






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Hint: With $$x=y-y^3$$ we get $$(y-y^3)^3-y+y^3+y=0$$ or $$(y-y^3)^3+y^3=0$$
                          factorizing we obtain
                          $$y^3(y^3-2)(y^4-y^2+1)=0$$ Can you proceed now?






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Hint: With $$x=y-y^3$$ we get $$(y-y^3)^3-y+y^3+y=0$$ or $$(y-y^3)^3+y^3=0$$
                            factorizing we obtain
                            $$y^3(y^3-2)(y^4-y^2+1)=0$$ Can you proceed now?






                            share|cite|improve this answer









                            $endgroup$



                            Hint: With $$x=y-y^3$$ we get $$(y-y^3)^3-y+y^3+y=0$$ or $$(y-y^3)^3+y^3=0$$
                            factorizing we obtain
                            $$y^3(y^3-2)(y^4-y^2+1)=0$$ Can you proceed now?







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 19 '18 at 19:09









                            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                            76.7k42866




                            76.7k42866























                                0












                                $begingroup$

                                Since $$x^3+y^3=0 implies (x+y)(x^2-xy+y^2)=0 $$



                                and $x^2-xy+y^2ne 0$ we have $x=-y$.



                                So $x^3-2x=0$ so $x=0$ (and $y=0$) or $x=pm sqrt{2} $ (and $y=mp sqrt{2} $ )






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  Since $$x^3+y^3=0 implies (x+y)(x^2-xy+y^2)=0 $$



                                  and $x^2-xy+y^2ne 0$ we have $x=-y$.



                                  So $x^3-2x=0$ so $x=0$ (and $y=0$) or $x=pm sqrt{2} $ (and $y=mp sqrt{2} $ )






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    Since $$x^3+y^3=0 implies (x+y)(x^2-xy+y^2)=0 $$



                                    and $x^2-xy+y^2ne 0$ we have $x=-y$.



                                    So $x^3-2x=0$ so $x=0$ (and $y=0$) or $x=pm sqrt{2} $ (and $y=mp sqrt{2} $ )






                                    share|cite|improve this answer









                                    $endgroup$



                                    Since $$x^3+y^3=0 implies (x+y)(x^2-xy+y^2)=0 $$



                                    and $x^2-xy+y^2ne 0$ we have $x=-y$.



                                    So $x^3-2x=0$ so $x=0$ (and $y=0$) or $x=pm sqrt{2} $ (and $y=mp sqrt{2} $ )







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Dec 19 '18 at 19:10









                                    greedoidgreedoid

                                    45.3k1159113




                                    45.3k1159113















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