algebra!

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x is a number such that x2+3x+9=0. What is the value of x3?

  • algebra! -

    x^2 + 3x + 9 = 0
    x = (-3 ±√-27)/2
    = (-3 ± 3i√3)/2

    if x = (-3 + 3i√3)/2
    x^3 = (1/8)( -27 + 3(9)(3i√3) + 3(-3)(3i√3)^2 + (3i√3)^3 )
    = (1/8)( -27 +81i√3 - 243i^2 + 81i^3√3)
    = (1/8)( -27 + 81i√3 + 243 - 81√3i)
    = (1/8)(216) = 27

    if x = (-3 - 3√3i)/2
    x^3 = (1/8)( -27 + 3(9)(-3√3i) +3(-3)(-3√3i)^2 + (-3√3)^3 )
    =(1/8)( -27 - 81√3i -243i^2√3 + (-3√3i)^3 )
    = (1/8)( -27 - 81√3i + 243 - 81√3i^3 )
    = (1/8) (-27 - 81√3i + 243 + 81√3i )
    =(1/8)(216) = 27

    so x^3 = 27

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