If $\alpha \text{\hspace{0.17em}and\hspace{0.17em}}\beta$ are the roots of the equation $x^2-x+1=0$, then the value of ${\alpha }^{2009}+{\beta }^{2009}$.

$x^2-x+1=0\Rightarrow x=\frac{1\pm \sqrt{1-4}}{2}$$x^2-x+1=0$

$x=\frac{1\pm \sqrt{3}i}{2}$

$\alpha =\frac{1}{2}+i\frac{\sqrt{3}}{2},\beta =\frac{1}{2}-i\frac{\sqrt{3}}{2}$

$=\cos \frac{\pi }{3}-i\sin \frac{\pi }{3}$

${\alpha }^{2009}+{\beta }^{2009}=2\cos 2009\left(\frac{\pi }{3}\right)$

$2\cos (668\pi +\pi +\frac{2\pi }{3})$

$=2\cos (\pi +\frac{2\pi }{3})$

$=-2\cos \frac{2\pi }{3}=-2\left(\frac{-1}{2}\right)=1$

(or)

Given $x^2-x+1=0$

$\therefore w_1-w^2$ are roots

$\therefore \alpha =-w,\beta =-w^2$

${(-w)}^{2009}+{(-w^2)}^{2009}=-w^{2009}-w^{4018}$

$=-w^{2009}-{\left(w^3\right)}^{1339}w$

$=-w^2-w$

$=-(w+w^2)$

$=1$