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