Explanation

(c)
Given, $\alpha,\ \beta$ are the roots of $x^{2} - 2x + 4 = 0$
$\therefore\ \alpha + \beta = 2$ ...(i)
And $\alpha\beta = 4$ ...(ii)
Now, $\alpha - \beta = \sqrt{(\alpha + \beta)^{2} - 4\alpha\beta}$
$= \sqrt{4 - 4 \times 4} = \sqrt{- 12}$
$\Rightarrow \ \alpha - \beta = 2\sqrt{3}i$ ...(iii)
On solving Eqs. (i) and (ii0, 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, $\alpha^{6} + \beta^{6} = \left( - 2\omega^{2} \right)^{6} + ( - 2\omega)^{6}$
$= 64\left( \omega^{3} \right)^{4} + 64\left( \omega^{3} \right)^{2}$
$= 128$ $\lbrack{\because\omega}^{3} = 1\rbrack$