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Different complex numbers

  1. Jan 4, 2013 #1
    We have a,b,c different complex numbers so

    (a+b)^3 = (b+c)^3 = (c+a)^3

    Show that a^3 = b^3 = c^3

    From the first equality I reached a^3 - c^3 + 3b(a-c)(a+b+c) = 0 How a is different from c => a-c is different from 0
    How do I show that a^3 - c^3 = 0?
  2. jcsd
  3. Jan 4, 2013 #2
    If you can prove that 3b(a-c)(a+b+c)=0, you're done.
    a, b, c are complex numbers.
  4. Jan 4, 2013 #3
    But how to prove that?
  5. Jan 4, 2013 #4
    Well, you can try assigning a value to the complex numbers and put them in the expression.
  6. Jan 4, 2013 #5
    Huh? How can that ever be a good proof??
  7. Jan 4, 2013 #6
    lol i think it can't be..but you can do it if you take the time..
    edit: expand the expression first.
    Last edited: Jan 4, 2013
  8. Jan 4, 2013 #7


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    That's a good start. (It will turn out that a+b+c=0.) If you take out the factor a-c, and also write down a corresponding equation with a/b/c rotated around, then add the two, what do you get?
  9. Jan 4, 2013 #8


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    Doh! There's a much easier way.
    We know (a+b) = (b+c)ω, where ω3=1. If a≠c then ω≠1. Writing the corresponding eqn for c+a versus b+c, and assuming b≠a and b≠c, we have (b+c) = (c+a)ω. (If (b+c) = (c+a)ω2 then b=c.) It's not hard from there.
  10. Jan 5, 2013 #9
    redount2k9, this formula can help you:


    You will need to prove 3b(a-c)(a+b+c)=0 using this forumla, as I said earlier. Or you can put ##a=x_1+iy_1, b=x_2+iy_2, c=x_3+iy_3##, but this will become lengthy.
  11. Jan 5, 2013 #10


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  12. Jan 5, 2013 #11
  13. Jan 5, 2013 #12
    I don't know if the original poster is still with us...

    We know a, b, and c are distinct.

    Therefore (a+b), (b+c) and (c+a) are also distinct

    Therefore , as per the question, they are the three distinct cube roots of some number z.

    And therefore what is the sum ((a+b) + (b+c) + (c+a))?
  14. Jan 5, 2013 #13
  15. Jan 5, 2013 #14
    More specifically, what is the sum of the three distinct cube roots of some complex number z?
  16. Jan 5, 2013 #15
    That's 0 ..there you go. This can be locked. :P
  17. Jan 5, 2013 #16
    Anyway. Use that if




    for [itex]k\in \mathbb{Z}[/itex]. And the same thing for the equality [itex](a+b)^3=(a+c)^3[/itex].

    Now solve the equations

    [tex]\left\{\begin{array}{l} a+b=e^{2ik\pi/3}(b+c)\\ a+b=e^{2ik^\prime\pi/3}(a+c) \end{array}\right.[/tex]
  18. Jan 5, 2013 #17
    Thread is open again. Let's actually wait for the OP to reply before helping any further :tongue2:
  19. Jan 6, 2013 #18
    What should I reply? All of you have different opinions... it's hard for me to understand something.
  20. Jan 6, 2013 #19
    hahahaha use any of the methods..
  21. Jan 6, 2013 #20
    Hahahaha why not to use the best method? (if I would know which it is)
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