Question

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

Answer 1

$x^3+1=\left(x+1\right)\left(x^2-x+1\right)$

So, since $\alpha$ is a root of $x^2-x+1=0$ we have,

$\begin{array}{c}{\alpha }^3+1=\left(\alpha +1\right)\left({\alpha }^2-\alpha +1\right)=\left(\alpha +1\right)\left(0\right)=0\\ So,\\ {\alpha }^3=-1\\ Similarly:\\ {\beta }^2=-1\\ Also:x^2-x+1=\left(x-\alpha \right)\left(x-\beta \right)=x^2-\left(\alpha +\beta \right)x+\alpha \beta \\ so:\{\begin{array}{c}\alpha +\beta =1\\ \alpha \beta =1\end{array}\\ {\alpha }^2+{\beta }^2={\left(\alpha +\beta \right)}^2-2\alpha \beta =1-2=-1\\ so,\\ {\alpha }^{2009}+{\beta }^{2009}={\alpha }^{3.669+2}+{\beta }^{3.669+2}\\ ={\left({\alpha }^3\right)}^{669}.{\alpha }^2+{\left({\beta }^3\right)}^{669}.{\beta }^2\\ ={\left(-1\right)}^{669}.\left({\alpha }^2+{\beta }^2\right)\\ =\left(-1\right)\left(-1\right)\\ =1\end{array}$