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If $\alpha, \beta$ are the roots of $1+x+x^2=0$, then the value of $\alpha^4+\beta^4+\alpha^{-4} \beta^{-4}$ is
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We have,
$$
\begin{aligned}
1+x+x^2 & =0 \\
\Rightarrow \quad x & =\frac{-1 \pm \sqrt{1-4}}{2} \Rightarrow x=\frac{-1 \pm \sqrt{3} i}{2} \\
\Rightarrow \quad x & =\omega, \omega^2
\end{aligned}
$$
It is given that, $\alpha, \beta$ are the roots of the equation $1+x+x^2=0$.
$$
\begin{aligned}
& \therefore \quad \alpha=\omega, \beta=\omega^2 \\
& \text { Now, } \alpha^4+\beta^4+\alpha^{-4} \beta^{-4}=\omega^4+\omega^8+\omega^{-4} \omega^{-8} \\
& =\omega+\omega^2+\omega^{-12} \\
& =-1+1=0 \quad\left[\because \omega^{-12}=\frac{1}{\omega^{12}}=1\right] \\
&
\end{aligned}
$$
$$
\begin{aligned}
1+x+x^2 & =0 \\
\Rightarrow \quad x & =\frac{-1 \pm \sqrt{1-4}}{2} \Rightarrow x=\frac{-1 \pm \sqrt{3} i}{2} \\
\Rightarrow \quad x & =\omega, \omega^2
\end{aligned}
$$
It is given that, $\alpha, \beta$ are the roots of the equation $1+x+x^2=0$.
$$
\begin{aligned}
& \therefore \quad \alpha=\omega, \beta=\omega^2 \\
& \text { Now, } \alpha^4+\beta^4+\alpha^{-4} \beta^{-4}=\omega^4+\omega^8+\omega^{-4} \omega^{-8} \\
& =\omega+\omega^2+\omega^{-12} \\
& =-1+1=0 \quad\left[\because \omega^{-12}=\frac{1}{\omega^{12}}=1\right] \\
&
\end{aligned}
$$
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