Ts Eamcet
Complex Numbers
TS EAMCET 2020 (Online) 11th September Evening Shift
MCQ+1 / -02020
If $z$ is a complex number such that $z^2+z+1=0$, then $\left(z+\frac{1}{z}\right)^3+\left(z^2+\frac{1}{z^2}\right)^3+\left(z^3+\frac{1}{z^3}\right)^3+\ldots . .+\left(z^{2020}+\frac{1}{z^{2020}}\right)^3=$