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Prove Equality given a condition

  1. Nov 20, 2007 #1

    Gib Z

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    1. The problem statement, all variables and given/known data
    If a+b+c = 0, show that [itex](2a-b)^3 + (2b-c)^3 + (2c-a)^3 = 3(2a-b)(2b-c)(2c-a)[/tex]


    2. Relevant equations

    None that I really need to state.

    3. The attempt at a solution

    Well I've just messed around with it, grinding through the algebra, and nothing seems to work. I did however notice that the (2a-b), (2b-c) and (2c-a) terms add up to a+b+c, and in this case, 0. So let u=(2a-b), v= (2b-c) and w=(2c-a) so that the question can be simplified to: if u+v+w = 0, then show [itex]u^3 + v^3 + w^3 = 3uvw[/itex].

    First I need verification if this change is correct, then hints on what I should do next =[
     
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  3. Nov 20, 2007 #2

    morphism

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    (u+v+w)^3 = 0

    Another hint:

    Of course (u+v+w)^3 is more profitably expressed as
    u^3+v^3+w^3+3u^2(v+w)+3v^2(u+w)+3w^2(u+v)+6uvw
     
  4. Nov 20, 2007 #3

    Gib Z

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    Yes I got that far with my messing around, but I ran into a wall when trying to see what to do with [itex]3u^2 (v+w) +3v^2 (u+w) +3w^2 (u+v) + 6uvw[/itex] :(
     
  5. Nov 20, 2007 #4

    rock.freak667

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    ah...this looks all roots of polynomial like...I shall help
    [tex](A+B)^3=A^3+3A^2B+3AB^2+B^3[/tex]

    now expand out (2a-b)^3 and (2b-c)^3 and (2c-a)^3 and simplify
    and then tell me what you get
     
  6. Nov 21, 2007 #5

    Gib Z

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    [tex](2a-b)^3 = 8a^3 - 4a^2b + 2ab^2 - b^3[/tex]

    The rest are the same, just replace the letters.

    [tex](2a-b)^3 +(2b-c)^3 +(2c-a)^3 = 7a^3 - 4a^2b + 2a^2c + 7b^3 - 4b^2c + 2b^a + 7c^3 - 4c^2a + 2bc^2[/tex]
     
  7. Nov 21, 2007 #6

    morphism

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    u+v+w=0 => u+v=-w, and so on.
     
  8. Nov 21, 2007 #7

    Gib Z

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    So I get that down to [tex]u^3+v^3+w^3 = 2uvw[/tex]. That is actually quite disturbing...
     
  9. Nov 21, 2007 #8

    Gib Z

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    Ahh sorry I forgot about the Original u^3+v^3+w^3 from post 2!! Thank you very much~!~!
     
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