If $\alpha$ and $\beta$ are the roots of the equation $x^2-x+1=0$, then $\alpha^{2009}+\beta^{2009}$ is equal to

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Aicharya · Accepted Answer

$\alpha$ and $\beta$ are the roots of quadratic equation $ \begin{aligned} x^2-x+1 & =0 \ x^2-x+1 & =0 \ x & =\frac{1 \pm \sqrt{1-4}}{2}=\frac{1 \pm \sqrt{-3}}{2} \ & =\frac{1 \pm \sqrt{3} i}{2}=-\omega 	ext { and }-\omega^2 \end{aligned} $ [where, $w$ is cube roots of unity] $	herefore-\omega,-\omega^2$ are roots of quadratic equation $ x^2-x+1=0 $ Let $\alpha=-w$, then $\beta=-\omega^2$ Now, $\alpha^{2009}+\beta^{2009}$ $ \begin{aligned} & =(-\omega)^{2009}+\left(-\omega^2ight)^{2009}=-\left[\omega^{3(669)+2}+\omega^{4018}ight] \ & =-\left[\left(\omega^3ight)^{669} \cdot \omega^2+\omega^{3(1339)+1}ight] \end{aligned} $ $\begin{aligned} & =-\left[\left(\omega^3ight)^{669} \cdot \omega^2+\left(\omega^3ight)^{1339} \cdot \omegaight] \ & {\left[\because \omega^3=1ight]} \ & =-\left[\omega^2+\omegaight] \ & {\left[\because 1+\omega+\omega^2=0ight]} \ & =-(-1)=1 \ & \end{aligned}$

If $\alpha$ and $\beta$ are the roots of the equation $x^2-x+1=0$, then $\alpha^{2009}+\beta^{2009}$ is…

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