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((-8)^2)^(1/3) <> (-8)^(2/3) <> (-8)^(1/3))^2

  1. Feb 10, 2016 #1
    ((-8)^2)^(1/3)=64^(1/3)=4
    (-8)^(1/3))^2 = (-2)^2=4
    but
    (-8)^(2/3) = 4(-1)^(2/3) = 4((2/3)(cos (π) + i*sin(π)) and so forth, giving a complex number.
    I understand why the latter one works. Why don't the first two?
    Thanks.
     
  2. jcsd
  3. Feb 10, 2016 #2
    Why does 2/3 act as a multiplicative constant instead of an exponential? Moreover ##\sin \pi =0##.
     
  4. Feb 10, 2016 #3

    fresh_42

    Staff: Mentor

    Your fifth equation isn't one.
     
  5. Feb 10, 2016 #4
    Oops, too quickly typed the last equation. I meant to multiply the two-thirds times the angles, not times the whole expression (or equivalently use it as an exponent), using Euler's formula. Anyway, my question is not about Euler's formula, which as I say I do get when I am not typing while my bus is about to leave, but rather why the first two formulas are invalid.
     
  6. Feb 10, 2016 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, it doesn't. sin(π)= 0. The other two cube roots are non-real, of course.
    The first two do also, you just have restricted yourself to real numbers. 64 has three complex cube roots, 4 is one, the other two are non-real. 8 has 3 cube roots, one is 2 and the other two are non-real.
     
  7. Feb 10, 2016 #6

    fresh_42

    Staff: Mentor

    They are valid. They give you one of possible three answers, the real one.
     
  8. Feb 11, 2016 #7
    Thanks, all of those who answered, everything clear now, and I apologize for the typos.
     
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