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