#### Answer

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

#### Work Step by Step

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