If a, b and c are cube roots of unity, then $\begin{vmatrix} e^{a} & e^{2a} & e^{3a}-1 \ e^{b} & e^{2b} & e^{3b}-1 \ e^{c} & e^{2c} & e^{3c}-1 \end{vmatrix}=$

Question

Accepted Answer

To solve the problem, we need to evaluate the determinant of the given matrix:

$$
\begin{vmatrix} 
e^{a} & e^{2a} & e^{3a}-1 \ 
e^{b} & e^{2b} & e^{3b}-1 \ 
e^{c} & e^{2c} & e^{3c}-1 
\end{vmatrix}
$$

where $a$, $b$, and $c$ are cube roots of unity. The cube roots of unity are $1$, $\omega$, and $\omega^2$, where $\omega = e^{2\pi i / 3}$ and $\omega^3 = 1$.

### Step-by-step Explanation:

1. **Properties of Cube Roots of Unity:**
   - The cube roots of unity satisfy the equation $x^3 = 1$, which implies $x^3 - 1 = 0$. Therefore, $x^3 = 1$ for each of the roots.
   - The sum of the cube roots of unity is zero: $1 + \omega + \omega^2 = 0$.

2. **Substitution in the Matrix:**
   - Substitute $a = 1$, $b = \omega$, and $c = \omega^2$ into the matrix:
     $$
     \begin{vmatrix} 
     e^{1} & e^{2} & e^{3}-1 \ 
     e^{\omega} & e^{2\omega} & e^{3\omega}-1 \ 
     e^{\omega^2} & e^{2\omega^2} & e^{3\omega^2}-1 
     \end{vmatrix}
     $$

3. **Simplification Using Properties:**
   - Since $e^{3} = e^{3 \cdot 1} = e^{3}$, $e^{3\omega} = e^{3 \cdot \omega} = e^{1} = e$, and $e^{3\omega^2} = e^{3 \cdot \omega^2} = e^{1} = e$, the matrix simplifies to:
     $$
     \begin{vmatrix} 
     e & e^{2} & e-1 \ 
     e^{\omega} & e^{2\omega} & e-1 \ 
     e^{\omega^2} & e^{2\omega^2} & e-1 
     \end{vmatrix}
     $$

4. **Determinant Evaluation:**
   - Notice that the third column becomes identical for all rows: $e - 1$.
   - This implies that the determinant of the matrix is zero because the columns are linearly dependent (i.e., the third column is a linear combination of the first two columns).

### Conclusion:

- **Correct Answer: Option 1 (0):** The determinant evaluates to zero due to the linear dependence of the columns when substituting the cube roots of unity into the matrix.
- **Incorrectness of Other Options:**
  - **Option 2 (e), Option 3 ($e^2$), and Option 4 ($e^3$):** These options suggest non-zero values for the determinant, which contradicts the linear dependence observed in the matrix. Since the determinant is zero, these options are incorrect.

Thus, the correct answer is indeed Option 1, which is 0.