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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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Verified Answer
The correct answer is:
-2
Given, $z^2+z+1=0$
$$
\begin{aligned}
& \left(z+\frac{1}{z}\right)^3+\left(z^4+\frac{1}{z^4}\right)^3=\left(\frac{z^2+1}{z}\right)^3+\left(\frac{z^8+1}{z^4}\right)^3 \\
& =\left(-\frac{z}{z}\right)^3+\left(\frac{z^8+1}{z^4}\right)^3\left[\because z^2+z+1=0 \Rightarrow z^2+1=-z\right] \\
& =-1+\left(\frac{z^8+1}{z^4}\right)^3 \\
& \because \quad z^2+1=-z \text {, squaring both side, } \\
& z^4+1+2 z^2=z^2 \Rightarrow z^4+1=-z^2
\end{aligned}
$$
Again, squaring it $\Rightarrow z^8+1+2 z^4=z^4$
$$
\Rightarrow \quad z^8+1=-z^4
$$
Putting in Eq. (i), we get
$$
\begin{aligned}
& \left(z+\frac{1}{z}\right)^3+\left(z^4+\frac{1}{z^4}\right)^3=-1+\left(-\frac{z^4}{z^4}\right)^3 \\
& =-1+(-1)=-2
\end{aligned}
$$
$$
\begin{aligned}
& \left(z+\frac{1}{z}\right)^3+\left(z^4+\frac{1}{z^4}\right)^3=\left(\frac{z^2+1}{z}\right)^3+\left(\frac{z^8+1}{z^4}\right)^3 \\
& =\left(-\frac{z}{z}\right)^3+\left(\frac{z^8+1}{z^4}\right)^3\left[\because z^2+z+1=0 \Rightarrow z^2+1=-z\right] \\
& =-1+\left(\frac{z^8+1}{z^4}\right)^3 \\
& \because \quad z^2+1=-z \text {, squaring both side, } \\
& z^4+1+2 z^2=z^2 \Rightarrow z^4+1=-z^2
\end{aligned}
$$
Again, squaring it $\Rightarrow z^8+1+2 z^4=z^4$
$$
\Rightarrow \quad z^8+1=-z^4
$$
Putting in Eq. (i), we get
$$
\begin{aligned}
& \left(z+\frac{1}{z}\right)^3+\left(z^4+\frac{1}{z^4}\right)^3=-1+\left(-\frac{z^4}{z^4}\right)^3 \\
& =-1+(-1)=-2
\end{aligned}
$$
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