How would one go about analytically solving a system of non-linear equations of the form: a + b + c = 4 a 2 + b 2 + c 2 = 6 a 3 + b 3 + c 3 = 10Thank you so much!

Question

How would one go about analytically solving a system of non-linear equations of the form:  a  +  b  +  c  =  4      a    2    +      b    2    +      c    2    =  6      a    3    +      b    3    +      c    3    =  10Thank you so much!

Semaj Stark · Accepted Answer

Hint: Newton&#039;s identities.                    a        +        b        +        c                            =        4                            a        b        +        b        c        +        c        a        =                  1          2                (        (        a        +        b        +        c                  )          2                −        (                  a          2                +                  b          2                +                  c          2                )        )                            =                  1          2                (                  4          2                −        6        )        =        5                            a        b        c        =                  1          3                (        (        a        b        +        b        c        +        c        a        −                  a          2                −                  b          2                −                  c          2                )        (        a        +        b        +        c        )        +        (                  a          3                +                  b          3                +                  c          3                )        )                            =                  1          3                (        (        5        −        6        )        ⋅        4        +        10        )        =        2            Remember that   a  ,     b  ,     c are the three roots of a polynomial  P  (  t  )  =  (  t  −  a  )  (  t  −  b  )  (  t  −  c  )  =      t    3    −  (  a  +  b  +  c  )      t    2    +  (  a  b  +  b  c  +  c  a  )  t  −  a  b  c

Adelyn Rodriguez · Accepted Answer

One solution is 1,1,2 by inspection.If we rewrite it in   d  =  a  −  1  ,  e  =  b  −  1  ,  f  =  c  −  1, the equations become  d  +  e  +  f  =  1      d    2    +      e    2    +      f    2    =  1      d    3    +      e    3    +      f    3    =  1and all the variables, if real, must be in   [  −  1  ,  1  ]. I would expect six solutions from the product of the degrees and have found three. So it is natural to assume two variables are equal to make three more, but the first two equations then yield       2    3    ,      2    3    ,            −      1        3  , which does not satisfy the last.

How would one go about analytically solving a system of non-linear equations of the form: a

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