Given, $\alpha, \beta$ are the roots of $x^2-2 x+4=0$


Now,
$\begin{aligned}
\alpha-\beta & =\sqrt{(\alpha+\beta)^2-4 \alpha \beta} \\
& =\sqrt{4-4 \times 4}=\sqrt{-12}
\end{aligned}$

On solving Eqs. (i) and (ii), we get
$\alpha=\frac{2+2 \sqrt{3} i}{2}=-2\left(\frac{-1-\sqrt{3} i}{2}\right)=-2 \omega^2$
and $\beta=\frac{2-2 \sqrt{3} i}{2}=-2\left(\frac{-1+\sqrt{3} i}{2}\right)=-2 \omega$
Now,
$\begin{aligned}
\alpha^6+\beta^6 & =\left(-2 \omega^2\right)^6+(-2 \omega)^6 \\
& =64\left(\omega^3\right)^4+64\left(\omega^3\right)^2 \\
& =128 \quad\left[\because \omega^3=1\right]
\end{aligned}$