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A question

  1. Nov 6, 2007 #1
    For n a nonnegative integer, what (in terms of n) is the sum of the n-th powers of the roots of the polynomial x^6 - 1 ?
  2. jcsd
  3. Nov 6, 2007 #2


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    The only roots of
    x^6 - 1 = 0
    are +1, and -1.


    1^n + (-1)^n
    1 + (-1)^n
  4. Nov 6, 2007 #3
    There are four more actually.
  5. Nov 6, 2007 #4


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    Sorry, of course there are; I foolishly put the others down to "complex" roots. Hmm.

    The 6 [real] roots are:

    x = 1
    x = (-1)^(1/3)
    x = (-1)^(2/3)
    x = -1
    x = -(-1)^(1/3)
    x = -(-1)^(2/3)

    That's the hard part done.
  6. Nov 6, 2007 #5
    Still wrong. There are not six real roots. There are two real roots ( 1, and -1) and then 4 complex roots. Do you, and it really should be the OP, know how to find the compex roots of this equation? In other words do you know how to find all the 6th roots of unity(one)?
  7. Nov 7, 2007 #6

    Gib Z

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    Homework Helper

    I would tend to believe (-1)^(1/3) is not real :( He could have expressed them better yes, but that is still correct.
  8. Nov 7, 2007 #7


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    The "nth roots of unity" are equally spaced around the unit circle. In particular that means the sum of the nth roots of unity themselves add to 0, if n is even, or 1, if n is odd. The 6th roots of unity add to 0. Obviously, the squares of the 6 roots of unity satisfy x3- 1= 0 so the sum of their squares is 1. Their third powers satisfy x2-1= 0 so the sum of their cubes is 0. The fourth powers are just the third powers, 1 and -1, times the roots themselves so we get the square roots twice and so the sum is 2. It's not too hard to show that the sum of the nth powers of the roots is 0 if n is odd, n/2 if n is even.

    The sixth roots of unity are, of course:
    [tex]e^{\frac{0i}{6}}= 1[/tex]
    [tex]e^{\frac{2\pi i}{6}}= \frac{1}{2}+\frac{\sqrt{3}}{2}[/tex]
    [tex]e^{\frac{4\pi i}{6}}= \frac{-1}{2}+\frac{\sqrt{3}}{2}[/tex]
    [tex]e^{\frac{6\pi i}{6}}= -1[/tex]
    [tex]e^{\frac{8\pi i}{6}}= \frac{-1}{2}-\frac{\sqrt{3}}{2}[/tex]
    [tex]e^{\frac{10\pi i}{6}= \frac{1}{2}-\frac{\sqrt{3}}{2}[/tex]
  9. Nov 7, 2007 #8
    I think that one is a bit debatable as all real numbers have a unique real cube root, but it may be that the principal cube root is probably complex, however this does not change the fact that he said those were the six real roots.
  10. Nov 7, 2007 #9
    They don't sum to 1 if n is odd...
    They sum to 1 if n is 1, and to 0 otherwise. Try it with n=3.
    It's easy to convince yourself about the rest by drawing a picture.
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