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# If $\alpha,\beta,\gamma$ are the cube roots of unity,then the value of the determinant $\begin{vmatrix}e^{\alpha}&e^{2\alpha}&e^{3\alpha}-1\\e^{\beta}&e^{2\beta}&e^{3\beta}-1\\e^{\gamma}&e^{2\gamma}&e^{3\gamma}-1\end{vmatrix}$ is equal to

$(a)\;-2\qquad(b)\;-1\qquad(c)\;0\qquad(d)\;1$

Given : $\alpha,\beta$ and $\gamma$ are the cube roots of unity,then assume
$\alpha=1,\beta=\omega$ and $v=\omega^2$
$\therefore\begin{vmatrix}e^\alpha&e^{2\alpha}&e^{3\alpha}-1\\e^\beta&e^{2\beta}&e^{3\beta}-1\\e^\gamma&e^{2\gamma}&e^{3\gamma}-1\end{vmatrix}=\begin{vmatrix}e^\alpha&e^{2\alpha}&e^{3\alpha}\\e^\beta&e^{2\beta}&e^{3\beta}\\e^\gamma&e^{2\gamma}&e^{3\gamma}\end{vmatrix}\begin{vmatrix}e^\alpha&e^{2\alpha}&-1\\e^\beta&e^{2\beta}&-1\\e^\gamma&e^{2\gamma}&-1\end{vmatrix}$
$\Rightarrow e^{\alpha}e^{\beta}e^{\gamma}\begin{vmatrix}1&e^{\alpha}&e^{2\alpha}\\1&e^{\beta}&e^{2\beta}\\1&e^{\alpha}&e^{2\gamma}\end{vmatrix}-\begin{vmatrix}1 &e^{\alpha}&e^{2\alpha}\\1&e^{\beta}&e^{2\beta}\\1&e^{\gamma}&e^{2\gamma}\end{vmatrix}$
$\Rightarrow \begin{vmatrix}1&e^{\alpha}&e^{2\alpha}\\1&e^{\beta}&e^{2\beta}\\1&e^{\gamma}&e^{2\gamma}\end{vmatrix}[e^{\alpha}e^{\beta}e^{\gamma}-1]=0$
Since $e^{\alpha}e^{\beta}e^{\gamma}=e^{1+\omega+\omega^2}$
$\Rightarrow e^0=1$
Hence (d) is the correct answer.
edited Mar 20, 2014