If $z$ is a complex number such that $z^2+z+1=0$,then $\left(z+\frac{1}{z}\right)^3+\left(z^2+\frac{1}{z^2}\right)^3+\left(z^3+\frac{1}{z^3}\right)^3+\ldots+\left(z^{2020}+\frac{1}{z^{2020}}\right)^3=$

If $\omega$ represents a complex cube root of unity,then $\left(1+\frac{1}{\omega}\right)\left(1+\frac{1}{\omega^2}\right)+\left(2+\frac{1}{\omega}\right)\left(2+\frac{1}{\omega^2}\right)+\ldots+\left(n+\frac{1}{\omega}\right)\left(n+\frac{1}{\omega^2}\right)=$

If $x=a+b$,$y=a \alpha+b \beta$,$z=a \beta+b \alpha$ and $\alpha, \beta$ are the complex cube roots of unity,then $x^3+y^3+z^3=$

If $\omega$ is a complex cube root of unity,then $225+(3 \omega+8 \omega^2)^2+(3 \omega^2+8 \omega)^2$ is equal to:

The product of the four values of the complex number $(1+i)^{3/4}$ is

If $1, \omega, \omega^2$ are the three cube roots of unity,then the value of $(a + b\omega + c\omega^2)^3 + (a + b\omega^2 + c\omega)^3$ is equal to,given that $a + b + c = 0$.