If `alpha, beta, gamma` are cube roots of unity, then the value of `|(e^(alpha),e^(2alpha),(e^(3alpha)-1)),(e^(beta),e^(2beta),(e^(3beta)-1)),(e^(gamma),e^(2gamma),(e^(3gamma)-1))|=`

Options

A. `-2`
B. `-1`
C. 0
D. 1

Answer

To solve the problem, we need to evaluate the determinant:

\[ D = \begin{vmatrix} e^{\alpha} & e^{2\alpha} & e^{3\alpha} - 1 \\ e^{\beta} & e^{2\beta} & e^{3\beta} - 1 \\ e^{\gamma} & e^{2\gamma} & e^{3\gamma} - 1 \end{vmatrix} \]

where \(\alpha, \beta, \gamma\) are the cube roots of unity.

### Step 1: Properties of Cube Roots of Unity
The cube roots of unity are given by:
\[ \alpha = 1, \quad \beta = e^{2\pi i / 3}, \quad \gamma = e^{4\pi i / 3} \]
These satisfy the properties:
\[ \alpha + \beta + \gamma = 0 \quad \text{and} \quad \alpha \beta \gamma = 1 \]

### Step 2: Rewrite the Determinant
We can rewrite the third column of the determinant:
\[ e^{3\alpha} - 1 = e^{3\alpha} - e^0 = e^{3\alpha} - e^{0} \]
This allows us to express the determinant as:
\[ D = \begin{vmatrix} e^{\alpha} & e^{2\alpha} & e^{3\alpha} \\ e^{\beta} & e^{2\beta} & e^{3\beta} \\ e^{\gamma} & e^{2\gamma} & e^{3\gamma} \end{vmatrix} - \begin{vmatrix} e^{\alpha} & e^{2\alpha} & 1 \\ e^{\beta} & e^{2\beta} & 1 \\ e^{\gamma} & e^{2\gamma} & 1 \end{vmatrix} \]

### Step 3: Factor Out Common Terms
Notice that \(e^{\alpha}, e^{\beta}, e^{\gamma}\) can be factored out from the first determinant:
\[ D = e^{\alpha + \beta + \gamma} \begin{vmatrix} 1 & e^{\alpha} & e^{2\alpha} \\ 1 & e^{\beta} & e^{2\beta} \\ 1 & e^{\gamma} & e^{2\gamma} \end{vmatrix} - \begin{vmatrix} e^{\alpha} & e^{2\alpha} & 1 \\ e^{\beta} & e^{2\beta} & 1 \\ e^{\gamma} & e^{2\gamma} & 1 \end{vmatrix} \]

### Step 4: Evaluate the Determinants
Since \(\alpha + \beta + \gamma = 0\), we have \(e^{\alpha + \beta + \gamma} = e^0 = 1\). Thus, we can simplify:
\[ D = \begin{vmatrix} 1 & e^{\alpha} & e^{2\alpha} \\ 1 & e^{\beta} & e^{2\beta} \\ 1 & e^{\gamma} & e^{2\gamma} \end{vmatrix} - \begin{vmatrix} e^{\alpha} & e^{2\alpha} & 1 \\ e^{\beta} & e^{2\beta} & 1 \\ e^{\gamma} & e^{2\gamma} & 1 \end{vmatrix} \]

### Step 5: Recognize the Determinants are Equal
The two determinants are equal due to the symmetry of the cube roots of unity. Therefore:
\[ D = 0 \]

### Final Result
Thus, the value of the determinant is:
\[ \boxed{0} \]