Help me with this system of equations a + b = 3 − c 1 a + 1 b = 5 12 − 1 c a 3 + b 3 = 45 − c 3

Question

Help me with this system of equations  a  +  b  =  3  −  c      1    a    +      1    b    =      5    12    −      1    c        a    3    +      b    3    =  45  −      c    3

Gerardo Barry · Accepted Answer

We have.                    a        +        b        +        c                            =        3                                      1          a                +                  1          b                +                  1          c                                    =                  5          12                                              a          3                +                  b          3                +                  c          3                                    =        45            The second equation tells us that             a      b      +      b      c      +      c      a              a      b      c        =      5    12  , or that   12  (  a  b  +  b  c  +  c  a  )  =  5  (  a  b  c  ).Using  (  a  +  b  +  c      )    3    −      a    3    −      b    3    −      c    3    =  3  (  a  +  b  )  (  b  +  c  )  (  c  +  a  )  =  3  (  3  −  a  )  (  3  −  b  )  (  3  −  c  ), we get that  27  −  45  =  3  [  27  −  9  (  a  +  b  +  c  )  +  3  (  a  b  +  b  c  +  c  a  )  −  a  b  c  ], or that  −  6  =  3  (  a  b  +  b  c  +  c  a  )  −  a  b  c  .Hence, we get that   a  b  +  b  c  +  c  a  =  −  10,   a  b  c  =  −  24. by solving the simultaneous equations.Hence, using Vieta, we get that   a  ,  b  ,  c are (permutation of) roots to the cubic      x    3    −  3      x    2    −  10  x  +  24  =  0.This cubic is   (  x  −  2  )  (  x  −  4  )  (  x  +  3  )  =  0, and thus we have   {  a  ,  b  ,  c  }  =  {  2  ,  4  ,  −  3  }.

Help me with this system of equations a + b = 3 − c

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