Suppose that real numbers a, b and c satisfy the condition a + b + c = 0. Prove the following identities: a 5 + b 5 + c 5 5 = a 3 + b 3 + c 3 3 ⋅ a 2 + b 2 + c 2 2 , a 7 + b 7 + c 7 7 = a 5 + b 5 + c 5 5 ⋅ a 2 + b 2 + c 2 2 .

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Suppose that real numbers a, b and c satisfy the condition  a  +  b  +  c  =  0. Prove the following identities:                    a        5            +              b        5            +              c        5              5    =                    a        3            +              b        3            +              c        3              3    ⋅                    a        2            +              b        2            +              c        2              2    ,                      a        7            +              b        7            +              c        7              7    =                    a        5            +              b        5            +              c        5              5    ⋅                    a        2            +              b        2            +              c        2              2    .

rotgelb7kjxw · Accepted Answer

This is a problem about elementary symmetric functions. Let      s    1    =  a  =  b  +  c      s    2    =  a  b  +  b  c  +  a  c      s    3    =  a  b  cNow assuming that       s    1    =  a  +  b  +  c  =  0 we show that                    −                                            a              2                        +                          b              2                        +                          c              2                                2                                    =                  s          2                =        a        b        +        b        c        +        c        a                                                                a              3                        +                          b              3                        +                          c              3                                3                                    =                  s          3                =        a        b        c                            −                                            a              4                        +                          b              4                        +                          c              4                                4                                    =        −                  1          2                          s          2          2                =        −                              (                          a              2                        +                          b              2                        +                          c              2                                      )              2                                2                                                                        a              5                        +                          b              5                        +                          c              5                                5                                    =        −                  s          2                          s          3                                    −                                            a              6                        +                          b              6                        +                          c              6                                6                                    =        −                  1          2                          s          3          2                +                  1          3                          s          2          3                                    −                                            a              7                        +                          b              7                        +                          c              7                                7                                    =                  s          2          2                          s          3                    And the results follow.The calculational scheme for deriving the above equalities isto note that  ln  ⁡  (  1  +  a  x  )  =  a  x  −            a      2        2        x    2    +            a      3        3        x    3    +  ⋯and adding,  ln  ⁡  (  1  +  a  x  )  (  1  +  b  x  )  (  1  +  c  x  )  =  (  a  +  b  +  c  )  x  −                    a        2            +              b        2            +              c        2              2        x    2    +                    a        3            +              b        3            +              c        3              3        x    3    +  ⋯on the other hand,  (  1  +  a  x  )  (  1  +  b  x  )  (  1  +  c  x  )  =  1  +      s    1    x  +      s    2        x    2    +      s    3        x    3  from which the above equalities follow easily by equating the coefficients of the two power series.In particular this shows that every                     a        n            +              b        n            +              c        n              n   is a polynomial in   n  =  2  ,  3 assuming       s    1    =  0

Suppose that real numbers a, b and c satisfy the condition a + b + c = 0 . Prove the f

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