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December 20, 2014

December 20, 2014

Posted by **please help me out!!! please** on Monday, February 11, 2013 at 10:50am.

- algebra! -
**Reiny**, Monday, February 11, 2013 at 11:17amx^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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