Good night!

[tex]\omega[/tex] is the imaginary cube root of unity.

All the numbers that are solution to:

[tex]x^3-1=0[/tex]

So:

[tex]x^3-1=(x-1)(x^2+x+1)=0[/tex]

Solving:

[tex]x=1[/tex]

This is one solution. There are another 2:

[tex]x^2+x+1=0\\

\Delta=(1)^2-4(1)(1)=-3\\

x=\dfrac{-1\pm\sqrt{-3}}{2}\\

x=\dfrac{-1\pm i\sqrt{3}}{2}\\

x'=\dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2}\\

x''=\dfrac{-1}{2}-i\dfrac{\sqrt{3}}{2}[/tex]

If [tex]\omega=\dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2}[/tex], so

[tex]\omega^2=\left(\dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2}\right)^2=\left(\dfrac{-1}{2}\right)^2+2\cdot\dfrac{-1}{2}\cdot i\dfrac{\sqrt{3}}{2}+\left(i\dfrac{\sqrt{3}}{2}\right)^2\\

\omega^2=\dfrac{1}{4}-i\dfrac{\sqrt{3}}{2}-\dfrac{3}{4}=\dfrac{-1}{2}-i\dfrac{\sqrt{3}}{2}[/tex]

So, the solutions are:

1, [tex]\omega[/tex] and [tex]\omega^2[/tex]

[tex]\omega^2+\omega+1=\dfrac{-1}{2}-i\dfrac{\sqrt{3}}{2}+\dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2}+1=-1+1=0[/tex]

I hope to have helped!