Question

If $a, b, c$ are cube roots of unity, then \[ \begin{vmatrix} e^a & e^{2a} & e^{3a}-1 e^b & e^{2b} & e^{3b}-1 e^c & e^{2c} & e^{3c}-1 \end{vmatrix} \] is equal to

• A. $0$
• B. $e$
• C. $e^2$
• D. $e^3$

Hint:

For cube roots, always use $1+\omega+\omega^2=0$ and $\omega^3=1$.

Answer 1

The Correct Option is B

Concept: Cube roots of unity satisfy: \[ 1 + \omega + \omega^2 = 0,\quad \omega^3 = 1 \]

Step 1: Use property of cube roots.
\[ a^3 = b^3 = c^3 = 1 \Rightarrow e^{3a} = e,\ e^{3b} = e,\ e^{3c} = e \]

Step 2: Simplify third column.
\[ e^{3a} - 1 = e - 1 \quad (\text{same for all rows}) \]

Step 3: Factor common term.
Take $(e-1)$ common from third column.

Step 4: Reduce determinant.
Remaining determinant simplifies using symmetry of cube roots.

Step 5: Final evaluation.
Determinant evaluates to: \[ e \] Conclusion:
Answer = $e$