If z^2+z+1=0,where z is a complex number,then | vedclass

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(D) Given,$z^2+z+1=0$. Since $z^2+z+1=0$,we have $z^2+1=-z$. Dividing by $z$,we get $z+\frac{1}{z}=-1$. Thus,$\left(z+\frac{1}{z}\right)^3 = (-1)^3 =

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-$2$

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(D) Given,$z^2+z+1=0$. Since $z^2+z+1=0$,we have $z^2+1=-z$. Dividing by $z$,we get $z+\frac{1}{z}=-1$. Thus,$\left(z+\frac{1}{z}\right)^3 = (-1)^3 = -1$. Also,$z^3=1$ (since $z$ is a cube root of unity). Then $z^4 = z^3 \cdot z = z$. So,$z^4+\frac{1}{z^4} = z+\frac{1}{z} = -1$. Therefore,$\left(z^4+\frac{1}{z^4}\right)^3 = (-1)^3 = -1$. Adding these results,$\left(z+\frac{1}{z}\right)^3+\left(z^4+\frac{1}{z^4}\right)^3 = -1 + (-1) = -2$.

If $z^2+z+1=0$,where $z$ is a complex number,then $\left(z+\frac{1}{z}\right)^3+\left(z^4+\frac{1}{z^4}\right)^3$ is equal to

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