Answer
$3(\cos 60^{\circ}+i\sin 60^{\circ})$,$3(\cos 180^{\circ}+i\sin 180^{\circ})$,$3(\cos 300^{\circ}+i\sin 300^{\circ})$
or
$[3{cis 60^{\circ},3cis 180^{\circ},3cis 300^{\circ}}]$
Work Step by Step
Given: $x^{3}+27=0$ or $x^{3}=-27$
This can be written in trigonometric form as: $x^{3}=27(\cos 180^{\circ}+i \sin 180^{\circ})$ $x=[27(\cos 180^{\circ}+i \sin 180^{\circ})]^{\frac{1}{3}}$
$x=3[(\cos 180^{\circ}+i \sin 180^{\circ})]^{\frac{1}{3}}$ $x=3[(\cos (180^{\circ}+2k\pi)+i \sin (180^{\circ}+2k\pi))]^{\frac{1}{3}}$
Apply De-Moivre's Theorem $x=3[(\cos \frac{(180^{\circ}+2k\pi)}{3}+i \sin\frac{(180^{\circ}+2k\pi)}{3})]$
Now, the arguments can be written as: $\frac{180^{\circ}+2k\pi}{3} $ and $k=0,1,2$
Roots: $3(\cos 60^{\circ}+i\sin 60^{\circ})$,$3(\cos 180^{\circ}+i\sin 180^{\circ})$,$3(\cos 300^{\circ}+i\sin 300^{\circ})$
Solution set of the equation can be written as: $[3{cis 60^{\circ},3cis 180^{\circ},3cis 300^{\circ}}]$