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Show that -1/2+ 1/2(i3 root) is a cube root of 1

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    Did you mean

    show that -1/2 + (1/2)√3 i is a cube root of 1 ??

    If so, then show that (-1/2 + (1/2)√3 i )^3 = 1

    LS = [ (-1/2) ( 1 - √3i) ]^3
    = (-1/2)^3 [ 1 + 3(-√3i) + 3(-√3i)^2 + (-√3i)^3 ]
    = (-18) [ 1 - 3√3i + 9i^2 - 3√3 i^3 ]
    = (-1/8)(1 - 3√3 - 9 + 3√3i ]
    =(-1/8)[-8}
    = 1
    = RS

    Yes, -1/2 + (1/2)√3 i is a cube root of 1

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