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!