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Math Help - Roots of a cubic polynomial

  1. #1
    Member jacs's Avatar
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    Roots of a cubic polynomial

    Looking for a quick identity to deal with the sum of the cubes of the roots of a cubic.

    I know that a + b + c = (a + b + c) - 2(ab + ac + bc)

    and i am trying to figure out one for

    a + b + c


    my teacher has it as a + b + c = (a + b + c) - 3(ab + ac + bc)

    but when i tried expanding the RHS i got
    a + b + c + 3ab + 3ac +3ab + 3ac + 3bc + 3bc - 3ab - 3ac - 3bc + 6abc
    which isnt even close to the identity we are after

    I have tried to come up it but haven't had any luck yet.

    thanks for any help
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  2. #2
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    Hello, jacs!

    Looking for a quick identity to deal with the sum of the cubes of the roots of a cubic.

    I know that: . a^2 + b^2 + c^2 \;=\; (a + b + c)^2 - 2(ab + ac + bc)

    and i am trying to figure out one for: . a^3 + b^3 + c^3

    My teacher has it as: . a^3 + b^3 + c^3 \:=\:(a + b + c)^3 - 3(ab + ac + bc) . . . . Wrong!

    \text{We have: }\;(a+b+c)^3 \;=\;a^3+b^3+c^3+3a^2b+3ab^2+ 3b^2c+3bc^2+3a^2c+3ac^2+6abc
    . . . . . . . . . . . . . . . . . . . . . . . . \underbrace{\qquad\qquad\qquad\qquad\qquad\qquad\q  quad\qquad\qquad\qquad}
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ^{\text{All of this is "extra"}}

    \text{Therefore: }\;a^3+b^3+c^3 \;=\;(a+b+c)^3\; - 3(a^2b+ab^2 +b^2c+bc^2+a^2c+ac^2+2abc)

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  3. #3
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    Quote Originally Posted by jacs View Post
    Looking for a quick identity to deal with the sum of the cubes of the roots of a cubic.

    I know that a + b + c = (a + b + c) - 2(ab + ac + bc)

    and i am trying to figure out one for

    a + b + c


    my teacher has it as a + b + c = (a + b + c) - 3(ab + ac + bc)

    but when i tried expanding the RHS i got
    a + b + c + 3ab + 3ac +3ab + 3ac + 3bc + 3bc - 3ab - 3ac - 3bc + 6abc
    which isnt even close to the identity we are after

    I have tried to come up it but haven't had any luck yet.

    thanks for any help
    Why not write  a^3 + b^3 + c^3 = (a+b+c)^3 + f(a,b,c)

    Expand  (a + b + c)^3 and see what is missing, and whatever is missing is what  f(a,b,c) is.

     (a+b+c)^3 = a^3 +b^3 + c^3+ 3a^2b + 3a^2c + 3ab^2 + 6abc + 3ac^2 + 3b^2c + 3bc^2

    Hence  f(a,b,c) = - 3a^2b - 3a^2c - 3ab^2 + 6abc - 3ac^2 - 3b^2c - 3bc^2

     =  -3(a^2b+a^2c +ab^2 + ac^2 + b^2c + bc^2 +3abc )

    and  a^3 + b^3 + c^3 = (a+b+c)^3 -3(a^2b+a^2c +ab^2 + ac^2 + b^2c + bc^2 +2abc )
    Last edited by Mush; April 3rd 2009 at 04:47 AM.
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  4. #4
    Member jacs's Avatar
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    thanks to both of you for that, i have gotten that far, but what i forgot to add in my original post was that i need to break it down into terms of
    a + b + c or
    ab + ac + bc or
    abc

    as we are working with sum and products of roots of a cubic

    from the looks of it though, there isnt going to be a nice easy fix
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  5. #5
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    Quote Originally Posted by jacs View Post
    thanks to both of you for that, i have gotten that far, but what i forgot to add in my original post was that i need to break it down into terms of
    a + b + c or
    ab + ac + bc or
    abc

    as we are working with sum and products of roots of a cubic

    from the looks of it though, there isnt going to be a nice easy fix
     a^3 + b^3 + c^3 = (a+b+c)^3 -3(a^2b+a^2c +ab^2 + ac^2 + b^2c + bc^2 +2abc )

      = (a+b+c)^3 -3(a(ab+ac) + b(ab +bc) + c(ac+cb) +2abc)
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