Answer :
Let [tex]\omega=e^{i\theta}[/tex]. By DeMoivre's theorem, we have
[tex]\omega^3=e^{3i\theta}=1\implies3i\theta=\ln1=0+2ni\pi\implies\theta=\dfrac{2n\pi}3[/tex]
where [tex]n[/tex] is any integer, but we can capture all the cube roots of unity by selecting [tex]n=0,1,2[/tex], and in particular when [tex]n=0[/tex] we get 1.
[tex]n=1\implies\omega=\dfrac{2i\pi}3=\cos\dfrac{2\pi}3+i\sin\dfrac{2\pi}3[/tex]
[tex]n=2\implies\omega^2=e^{4i\pi/3}=\cos\dfrac{4\pi}3+i\sin\dfrac{4\pi}3[/tex]
Recall that [tex]\cos(2\pi-x)=\cos x[/tex], so that [tex]\cos\dfrac{4\pi}3=\cos\dfrac{2\pi}3[/tex]. On the other hand, [tex]\sin(2\pi-x)=-\sin x[/tex], so [tex]\sin\dfrac{4\pi}3=-\sin\dfrac{2\pi}3[/tex]. So in [tex]\omega+\omega^2[/tex], we find the imaginary parts cancel, leaving us with
[tex]1+\omega+\omega^2=1+2\mathrm{Re}(\omega)[/tex]
for which we have
[tex]\mathrm{Re}(\omega)=\cos\dfrac{2\pi}3=-\dfrac12[/tex]
so the sum is [tex]1+\omega+\omega^2=0[/tex].
Another way of computing the sum is to notice that
[tex](1-\omega)(1+\omega+\omega^2)=1-\omega^3[/tex]
for any [tex]\omega[/tex]. We assumed that [tex]\omega^3=1[/tex], so the RHS is 0. Then for [tex]\omega\neq1[/tex], we must have [tex]1+\omega+\omega^2=0[/tex].
[tex]\omega^3=e^{3i\theta}=1\implies3i\theta=\ln1=0+2ni\pi\implies\theta=\dfrac{2n\pi}3[/tex]
where [tex]n[/tex] is any integer, but we can capture all the cube roots of unity by selecting [tex]n=0,1,2[/tex], and in particular when [tex]n=0[/tex] we get 1.
[tex]n=1\implies\omega=\dfrac{2i\pi}3=\cos\dfrac{2\pi}3+i\sin\dfrac{2\pi}3[/tex]
[tex]n=2\implies\omega^2=e^{4i\pi/3}=\cos\dfrac{4\pi}3+i\sin\dfrac{4\pi}3[/tex]
Recall that [tex]\cos(2\pi-x)=\cos x[/tex], so that [tex]\cos\dfrac{4\pi}3=\cos\dfrac{2\pi}3[/tex]. On the other hand, [tex]\sin(2\pi-x)=-\sin x[/tex], so [tex]\sin\dfrac{4\pi}3=-\sin\dfrac{2\pi}3[/tex]. So in [tex]\omega+\omega^2[/tex], we find the imaginary parts cancel, leaving us with
[tex]1+\omega+\omega^2=1+2\mathrm{Re}(\omega)[/tex]
for which we have
[tex]\mathrm{Re}(\omega)=\cos\dfrac{2\pi}3=-\dfrac12[/tex]
so the sum is [tex]1+\omega+\omega^2=0[/tex].
Another way of computing the sum is to notice that
[tex](1-\omega)(1+\omega+\omega^2)=1-\omega^3[/tex]
for any [tex]\omega[/tex]. We assumed that [tex]\omega^3=1[/tex], so the RHS is 0. Then for [tex]\omega\neq1[/tex], we must have [tex]1+\omega+\omega^2=0[/tex].