If z^2+z+1=0,then find the value of left(z^3+ | vedclass

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(C) Given that $z$ is a complex cube root of unity,we have $z^3 = 1$ and $1+z+z^2 = 0$.From $1+z+z^2 = 0$,we get $z^2+1 = -z$.Now,consider the express

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$5$

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(C) Given that $z$ is a complex cube root of unity,we have $z^3 = 1$ and $1+z+z^2 = 0$.From $1+z+z^2 = 0$,we get $z^2+1 = -z$.Now,consider the expression $\left(z^3+\frac{1}{z^3}\right)^2+\left(z^4+\frac{1}{z^4}\right)^2$.Substituting $z^3 = 1$,we get $\left(1+\frac{1}{1}\right)^2+\left(z^3 \cdot z+\frac{1}{z^3 \cdot z}\right)^2$.$= (1+1)^2 + \left(z+\frac{1}{z}\right)^2$.$= 4 + \left(\frac{z^2+1}{z}\right)^2$.Substituting $z^2+1 = -z$,we get $4 + \left(\frac{-z}{z}\right)^2$.$= 4 + (-1)^2 = 4 + 1 = 5$.

If $z^2+z+1=0$,then find the value of $\left(z^3+\frac{1}{z^3}\right)^2+\left(z^4+\frac{1}{z^4}\right)^2$,where $z$ is a complex cube root of unity.

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