If alpha and beta be the roots of equation x^2-x+1=0, then alpha^2009 + beta^2009 is equal to?          [Ans: 1]

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$$\begin{gathered} {\mathrm{x}}^2-\mathrm{x}+1=0 \\ \mathrm{By} \mathrm{Shridharacharya}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have} \\ \mathrm{x}=\frac{1\pm \sqrt{1-4}}{2}=\frac{1\pm \sqrt{-3}}{2}=\frac{1\pm \mathrm{i}\sqrt{3}}{2} \\ =\frac{1+\mathrm{i}\sqrt{3}}{2} \mathrm{or} \frac{1-\mathrm{i}\sqrt{3}}{2} \\ =-\left(\frac{-1-\mathrm{i}\sqrt{3}}{2}\right) \mathrm{or} -\left(\frac{-1+\mathrm{i}\sqrt{3}}{2}\right) \\ =-{\mathrm{\omega }}^2 \mathrm{or} -\mathrm{\omega } \\ \mathrm{Hence} \\ \mathrm{\alpha }=-\mathrm{\omega } \mathrm{and} \mathrm{\beta }=-{\mathrm{\omega }}^2 \\ \mathrm{Also} \mathrm{we} \mathrm{know} \mathrm{that} \\ {\mathrm{\omega }}^{3\mathrm{n}}=1 \mathrm{and} \\ 1+\mathrm{\omega }+{\mathrm{\omega }}^2=1..\left(\mathrm{i}\right) \\ {\mathrm{\alpha }}^{2009}+{\mathrm{\beta }}^{2009} \\ ={\mathrm{\alpha }}^{2007}.{\mathrm{\alpha }}^2+{\mathrm{\beta }}^{2007}.{\mathrm{\beta }}^2 \left[2007 \mathrm{is} \mathrm{a} \mathrm{multiple} \mathrm{of} 3\right] \\ ={\left(-\mathrm{\omega }\right)}^{2007} {\left(-\mathrm{\omega }\right)}^2+{\left(-{\mathrm{\omega }}^2\right)}^{2007} {\left(-{\mathrm{\omega }}^2\right)}^2 \\ =-{\left(\mathrm{\omega }\right)}^{2007} {\left(\mathrm{\omega }\right)}^2-{\left({\mathrm{\omega }}^2\right)}^{2007} \left({\mathrm{\omega }}^4\right) \\ =-1.{\left(\mathrm{\omega }\right)}^2-1.{\mathrm{\omega }}^3.\mathrm{\omega } \\ =-{\mathrm{\omega }}^2-\mathrm{\omega } \\ \mathrm{Using} \left(\mathrm{i}\right) \mathrm{we} \mathrm{get} \\ =1 \left[\mathrm{Answer}\right] \end{gathered}$$

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