#### Answer

$[3{cis 60^{\circ},3cis 180^{\circ},3cis 300^{\circ}}]$

#### 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
$x=3[(\cos (180^{\circ}+2k\pi)/3+i \sin (180^{\circ}+2k\pi)/3)]$
Now, the arguments can be wriiten as:
$\frac{180+2k\pi}{3} $ and $k=0,1,2$
Roots: $3(\cos 60^{\circ}+\sin 60^{\circ})$,$3(\cos 180^{\circ}+\sin 180^{\circ})$,$3(\cos 300^{\circ}+\sin 300^{\circ})$
Solution set of the equation can be written as:
$[3{cis 60^{\circ},3cis 180^{\circ},3cis 300^{\circ}}]$