Given equation is $x^{2}-x+1=0$
$$
\begin{aligned}
x &=\frac{1 \pm \sqrt{1-4}}{2} \quad\left(\because x=\frac{b \pm \sqrt{b^{2}-4 a c}}{2 a}\right) \\
&=\frac{1 \pm i \sqrt{3}}{2}=\frac{1+i \sqrt{3}}{2}, \frac{1-i \sqrt{3}}{2} \\
\Rightarrow-x &=\frac{-1+i \sqrt{3}}{2}, \frac{-1-i \sqrt{3}}{2} \\
\Rightarrow+x &=\omega,-\omega^{2}
\end{aligned}
$$
Since, $(\alpha, \beta)$ are the roots of given equation. Then, $\alpha=-\omega$ and $\beta=-\omega^{2}$
$$
\begin{aligned}
\therefore \alpha^{2013}+\beta^{2013} &=-(\omega)^{2013}+\left(-\omega^{2}\right)^{2013} \\
&=-\omega^{2013}-\omega^{4026} \\
&=-\left(\omega^{3}\right)^{671}-\left(\omega^{3}\right)^{1342} \\
&=-(1)^{671}-(1)^{1342} \\
&=-1-1=-2
\end{aligned}
$$