Let $\alpha$ and $\beta$ be the roots of the equation $x^{2}-x+1=0$.
$\alpha+\beta=-(-1)=1$...(i)
$\alpha \beta=1$
Now, $\alpha-\beta=\sqrt{(\alpha+\beta)^{2}-4 \alpha \beta}$
$=\sqrt{(1)^{2}-4(1)}=\sqrt{-3}=\sqrt{3} i$...(ii)
On solving(i), (ii) we get
$\alpha=\frac{1+i \sqrt{3}}{2}$ and $\beta=\frac{1-i \sqrt{3}}{2}$
$\Rightarrow \alpha=\frac{1}{2}+i \cdot \frac{\sqrt{3}}{2}$ and $\beta=\frac{1}{2}-i \cdot \frac{\sqrt{3}}{2}$
$\Rightarrow \quad \alpha=\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}$ and $\beta=\cos \frac{\pi}{3}-i \sin \frac{\pi}{3}$
(a) $\alpha^{4}-\beta^{4}=\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}-\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}$
$=2 i \sin \frac{4 \pi}{3}$
(By Demoiver's thm)
$\Rightarrow \alpha^{4}-\beta^{4}$ is not real.
(b) $2\left(\alpha^{5}+\beta^{5}\right)$
$=2\left(\cos \frac{5 \pi}{3}+i \sin \frac{5 \pi}{3}+\cos \frac{5 \pi}{3}-i \sin \frac{5 \pi}{3}\right)$
$=2.2 \cos \frac{5 \pi}{3}=4 \cdot \frac{1}{2}=2$
Now, $(\alpha \beta)^{5}=1$
$\Rightarrow 2\left(\alpha^{5}+\beta^{5}\right) \neq(\alpha \beta)^{5}$
(c) $\alpha^{6}-\beta^{6}=\cos \frac{6 \pi}{3}+i \sin \frac{6 \pi}{3}-\cos \frac{6 \pi}{3}+i \sin \frac{6 \pi}{3}$
$=2 i \sin 2 \pi=0$
Hence, option (c) is correct.