Use this formula to find all the solutions of the equation? $x^3$ + 8 =0

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Answer 1 · 2016-12-03T17:37:28

See below.

Using de Moivre's identity

$e^(i phi) = cosphi + i sin phi$

$x^3= -8 = -2^3=2^3e^(i(pi+2kpi))$ for $k=0,1,2$ then

$x = 2 e^(i(pi+2kpi)/3)$ so the roots are:

$2 e^(ipi/3), 2 e^(ipi),2 e^(i(5pi)/3)$ or

$1 + i sqrt[3], -2, 1 - i sqrt[3]$