#### Answer

$[2{cis 0^{\circ},2cis 120^{\circ},2cis 240^{\circ}}]$

#### Work Step by Step

Given: $x^{3}-8=0$
or $x^{3}=8$
This can be written in trigonometric form as:
$x^{3}=8+0.i=8(\cos 180^{\circ}+i \sin 180^{\circ})$
$x=[8(\cos 180^{\circ}+i \sin 180^{\circ})]^{\frac{1}{3}}$
$x=2[(\cos 180^{\circ}+i \sin 180^{\circ})]^{\frac{1}{3}}$
Absolute value of third root is given as $\sqrt[3] 1=1$
Now, the arguments can be wriiten as:
$k=0,1,2$
Roots: $2(\cos 0^{\circ}+\sin 0^{\circ})$,$2(\cos 120^{\circ}+\sin 120^{\circ})$,$2(\cos 240^{\circ}+\sin 240^{\circ})$
Solution set of the equation can be written as:
$[2{cis 0^{\circ},2cis 120^{\circ},2cis 240^{\circ}}]$