Explanation

(c)
Given, $\alpha,\beta\$and$\$γ are$\$the cube roots of unity, then assume
$\alpha = 1,\beta = \omega\$and$\ \gamma = \omega^{2}.$
$\therefore\left| \begin{matrix}
e^{\alpha} & e^{2\alpha} & (e^{3\alpha} - 1) \\
e^{\beta} & e^{2\beta} & (e^{3\beta} - 1) \\
e^{\gamma} & e^{2\gamma} & (e^{3\gamma} - 1)
\end{matrix} \right|$
$= \left| \begin{matrix}
e^{\alpha} & e^{2\alpha} & e^{3\alpha} \\
e^{\beta} & e^{2\beta} & e^{3\beta} \\
e^{\gamma} & e^{2\gamma} & e^{3\gamma}
\end{matrix} \right| + \left| \begin{matrix}
e^{\alpha} & e^{2\alpha} & - 1 \\
e^{\beta} & e^{2\beta} & - 1 \\
e^{\gamma} & e^{2\gamma} & - 1
\end{matrix} \right|$
$= e^{\alpha}e^{\beta}e^{\gamma}\left| \begin{matrix}
1 & e^{\alpha} & e^{2\alpha} \\
1 & e^{\beta} & e^{2\beta} \\
1 & e^{\gamma} & e^{2\gamma}
\end{matrix} \right| - \left| \begin{matrix}
1 & e^{\alpha} & e^{2\alpha} \\
1 & e^{\beta} & e^{2\beta} \\
1 & e^{\gamma} & e^{2\gamma}
\end{matrix} \right|$
$= \left| \begin{matrix}
1 & e^{\alpha} & e^{2\alpha} \\
1 & e^{\beta} & e^{2\beta} \\
1 & e^{\gamma} & e^{2\gamma}
\end{matrix} \right|$[$e^{\alpha}e^{\beta}e^{\gamma} - 1\rbrack = 0$
($\because e^{\alpha}e^{\beta}e^{\gamma} = e^{1 + \omega + \omega^{2}} = e^{0} = 1)$