If a x 2 + b x + c = 0 and b x 2 + c x + a = 0 have a common root and a ≠ 0, then find a 3 + b 3 + c 3 a b c I tried that for both equations to have a common root, the expression on left hand sides must be equal, ie a x 2 + b x + c = b x 2 + c x + afor this we must have x = 1 (i cannot prove this, but it appears to be true). Also both of these must be equal to 0, so we have: a + b + c = 0So using this we say a 3 + b 3 + c 3 a b c = a 3 + b 3 + c 3 − 3 a b c + 3 a b c a b c = ( a + b + c ) ( . . . ) a b c + 3So we get the answer as 3How do we say that x = 1 is the commmon root? Thanks!

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If   a      x    2    +  b  x  +  c  =  0 and   b      x    2    +  c  x  +  a  =  0 have a common root and   a  ≠  0, then find                     a        3            +              b        3            +              c        3                    a      b      c      I tried that for both equations to have a common root, the expression on left hand sides must be equal, ie  a      x    2    +  b  x  +  c  =  b      x    2    +  c  x  +  afor this we must have   x  =  1 (i cannot prove this, but it appears to be true). Also both of these must be equal to   0, so we have:  a  +  b  +  c  =  0So using this we say                    a        3            +              b        3            +              c        3                    a      b      c        =                    a        3            +              b        3            +              c        3            −      3      a      b      c      +      3      a      b      c              a      b      c        =            (      a      +      b      +      c      )      (      .      .      .      )              a      b      c        +  3So we get the answer as   3How do we say that   x  =  1 is the commmon root? Thanks!

beentjie8e · Accepted Answer

0  =  a      x    3    +  b      x    2    +  c  x  =  a      x    3    −  awhich gives   x  =  1,   a  +  b  +  c  =  0 and since      a    3    +      b    3    +      c    3    −  3  a  b  c  =  (  a  +  b  +  c  )  (      a    2    +      b    2    +      c    2    −  a  b  −  a  c  −  b  c  )  =  0  ,we obtain:                    a        3            +              b        3            +              c        3                    a      b      c        =            3      a      b      c              a      b      c        =  3.If you wish the solution for   x  ∈      C   and   {  a  ,  b  ,  c  }  ⊂      C   then since      a    3    +      b    3    +      c    3    −  3  a  b  c  =  (  a  +  b  +  c  )  (  a  +  ζ  b  +      ζ    2    c  )  (  a  +      ζ    2    b  +  ζ  c  )  ,where   ζ  =  −      1    2    +            3        2    i, we get the same answer.

Jenny Stafford · Accepted Answer

We have  f  (  x  )  =  a      x    2    +  b  x  +  c    g  (  x  )  =  b      x    2    +  c  x  +  a  f  (  x  )  =  a      x    2    +  b  x  +  c    g  (  x  )  =  b      x    2    +  c  x  +  aIf we set   t as the common root, then we know that   f  (  t  )  =  g  (  t  )  =  0                    a                  t          2                +        b        t        +        c                            =        b                  t          2                +        c        t        +        a            We can equate coefficents to conclude that   a  =  b  =  cTherefore, we can say that                                                        a              3                        +                          b              3                        +                          c              3                                            a            b            c                                              =                                            a              3                        +                          a              3                        +                          a              3                                            a            a            a                                                            =                              3                          a              3                                            a            3                                                            =        3

If ax^2+bx+c=0 and bx^2+cx+a=0 have a common root and a≠0, then find (a^3+b^3+c^3)/abc

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