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Math Help - Cube Roots Of Unity (Quadratic Equations)

  1. #1
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    Post Cube Roots Of Unity (Quadratic Equations)

    ,

    If 1, Ω, Ω^2 are the cube roots of unity, find the value of:

    1) Ω^2 + Ω^3 + Ω^4:

    Ans:

    Since Ω^3 = 1

    Ω^2 + 1 + Ω

    So, Ω^2 + Ω^3 + Ω^4 = 0

    I did not understand how the final answer result to be a 0.
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Cube roots of unity are the roots of x^3-1. Note that x^3-1 = (x-1)(x^2+x+1). Since \Omega is not a root of x-1 it must be a root of x^2+x+1.

    Now just notice that \Omega^2+\Omega^3+\Omega^4=\Omega^2(1+\Omega+\Omeg  a^2)=0.
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  3. #3
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    Quote Originally Posted by Bruno J. View Post
    Cube roots of unity are the roots of x^3-1. Note that x^3-1 = (x-1)(x^2+x+1). Since \Omega is not a root of x-1 it must be a root of x^2+x+1.

    Now just notice that \Omega^2+\Omega^3+\Omega^4=\Omega^2(1+\Omega+\Omeg  a^2)=0.
    This means that x^2+x+1 is always equal to 0 and hence (1+\Omega+\Omega^2) = 0, am i correct?
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    MHF Contributor Bruno J.'s Avatar
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    It's not always equal to zero, no. For instance it's equal to 3 when x=1. However it's equal to 0 when x is a cube root of unity, so \Omega^2+\Omega+1=0 is right!
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  5. #5
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    Quote Originally Posted by Bruno J. View Post
    It's not always equal to zero, no. For instance it's equal to 3 when x=1. However it's equal to 0 when x is a cube root of unity, so \Omega^2+\Omega+1=0 is right!
    I've understood it partially. I think i am unclear with the concept of "cube root of unity" If you can explain what it actually means, i would be grateful.
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