How to solve system equation { x 3 + y 3 + z 3 = x + y + z x 2 + y 2 + z 2 = x y z x , y , z ∈ R

Question

How to solve system equation      {                                        x            3                    +                      y            3                    +                      z            3                    =          x          +          y          +          z                                                          x            2                    +                      y            2                    +                      z            2                    =          x          y          z                                  x  ,  y  ,  z  ∈  R

trantegisis · Accepted Answer

We&#039;ll using the following notation:      s    1    =  x  +  y  +  z      s    2    =      x    2    +      y    2    +      z    2        s    3    =      x    3    +      y    3    +      z    3  And we&#039;ll express   x  ,  y  ,  z as root of the cubic equation  f  (  x  )  =  a      x    3    +  b      x    2    +  c  x  +  dFrom the Vieta&#039;s formulas we know that   d  =  −  x  y  zFrom the condition we have   d  =  −      s    2   and       s    1    =      s    3  Now use Newton&#039;s identitities and we have:  a      s    1    +  b  =  0  b  =  −  a      s    1  Now repeat the process:  a      s    2    +  b      s    1    +  2  c  =  0  a      s    2    +  −  a      s    1    2    +  2  c  =  0  a  (      s    2    −      s    1    2    )  +  2  c  =  0  2  c  =  −  a  (      s    2    −      s    1    2    )  c  =            −      a      (              s        2            −              s        1        2            )        2  And one last time:  a      s    3    +  b      s    2    +  c      s    1    +  3  d  =  0  a      s    1    −  a      s    1        s    2    +            −      a              s        1            (              s        2            −              s        1        2            )        2    −  3      s    2    =  0  a      s    1    −  a      s    1        s    2    +            −      a              s        1            (              s        2            −              s        1        2            )        2    −  3      s    2    =  0  a      s    1        (    1    −          s      2        −                            s          2                −                  s          1          2                    2        )    −  3      s    2    =  0  a      s    1        (                  −                  s          2                +                  s          1          2                +        2        −        2                  s          2                    2        )    =  3      s    2    a  =            6              s        2                            s        1            (      −              s        2            +              s        1        2            +      2      −      2              s        2            )      Now for arbitrary       s    1   and       s    2   you should be able to obtain a cubic equation, which roots will satisfy the condition.

uri2e4g · Accepted Answer

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