Question:
The value of the determinant
\[
\begin{vmatrix}
(a^x + a^{-x})^2 & (a^x - a^{-x})^2 & 1
(b^x + b^{-x})^2 & (b^x - b^{-x})^2 & 1
(c^x + c^{-x})^2 & (c^x - c^{-x})^2 & 1
\end{vmatrix}
\]
is
The value of the determinant
\[
\begin{vmatrix}
(a^x + a^{-x})^2 & (a^x - a^{-x})^2 & 1
(b^x + b^{-x})^2 & (b^x - b^{-x})^2 & 1
(c^x + c^{-x})^2 & (c^x - c^{-x})^2 & 1
\end{vmatrix}
\]
is
Show Hint
Use identities to simplify determinant columns before expanding.
Updated On: Apr 23, 2026
- $abc$
- $2abc$
- $a^2b^2c^2$
- None of these
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Concept:
Use identity:
\[
(a^x + a^{-x})^2 - (a^x - a^{-x})^2 = 4
\]
Step 1: Apply identity.
\[ (a^x + a^{-x})^2 = (a^x - a^{-x})^2 + 4 \]
Step 2: Column operation.
Make $C_1 = C_1 - C_2$ → constant column.
Step 3: Factorization.
Determinant reduces to product form.
Step 4: Final simplification.
\[ = abc \] Conclusion:
Answer = $abc$
Step 1: Apply identity.
\[ (a^x + a^{-x})^2 = (a^x - a^{-x})^2 + 4 \]
Step 2: Column operation.
Make $C_1 = C_1 - C_2$ → constant column.
Step 3: Factorization.
Determinant reduces to product form.
Step 4: Final simplification.
\[ = abc \] Conclusion:
Answer = $abc$
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