The cube roots of $$8$$ are plotted in the complex plane on the circle of radius $$2$$
They can be written as
$$2\cos 0+i\sin 0=2$$
$$2\left( \cos { \left( \cfrac { 2\pi }{ 3 } \right) } +i\sin { \left( \cfrac { 2\pi }{ 3 } \right) } \right) =-1+\sqrt { 3 } i=2\omega $$
$$2\left( \cos { \left( \cfrac { 4\pi }{ 3 } \right) } +i\sin { \left( \cfrac { 4\pi }{ 3 } \right) } \right) =-1-\sqrt { 3 } i=2{\omega}^{2} $$
find all the roots of $${x}^{3}-8=0$$
$${x}^{3}-8=(x-2)({x}^{2}+2x+4)$$
Quadratic formula:
$$x=\cfrac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } =\cfrac { -b\pm \sqrt { { 2 }^{ 2 }-(4\times 1\times 4) } }{ 2\cdot 1 } =-1\pm \sqrt { 3 } i$$