Question:
If \( x \) is a complex root of the equation
\[
\begin{vmatrix} 1 & x & x \\ x & 1 & x \\ x & x & 1 \end{vmatrix}
+
\begin{vmatrix} 1 - x & 1 & 1 \\ 1 & 1 - x & 1 \\ 1 & 1 & 1 - x \end{vmatrix} = 0,
\]
then \( x^{2007} + x^{-2007} \) is:
If \( x \) is a complex root of the equation
\[ \begin{vmatrix} 1 & x & x \\ x & 1 & x \\ x & x & 1 \end{vmatrix} + \begin{vmatrix} 1 - x & 1 & 1 \\ 1 & 1 - x & 1 \\ 1 & 1 & 1 - x \end{vmatrix} = 0, \] then \( x^{2007} + x^{-2007} \) is:
\[ \begin{vmatrix} 1 & x & x \\ x & 1 & x \\ x & x & 1 \end{vmatrix} + \begin{vmatrix} 1 - x & 1 & 1 \\ 1 & 1 - x & 1 \\ 1 & 1 & 1 - x \end{vmatrix} = 0, \] then \( x^{2007} + x^{-2007} \) is:
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If \( x \) is a cube root of unity, then it satisfies \( x^3 = -1 \), which helps simplify large exponents.
Updated On: Mar 26, 2025
- \( 1 \)
- \( -1 \)
- \( -2 \)
- \( 2 \)
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The Correct Option is C
Solution and Explanation
Step 1: {Expand both determinants}
\[ (1 - 3x^2 + 2x^3) + (3x^2 - x^3) = 0. \] Step 2: {Solve for \( x \)}
\[ x^3 + 1 = 0. \] \[ x^3 = -1. \] \[ x = -\omega, -\omega^2, -1. \] Step 3: {Compute \( x^{2007} + x^{-2007} \)}
Since \( x^3 = -1 \), \[ x^{2007} = (-1)^{669} = -1. \] \[ x^{-2007} = -1. \] Step 4: {Conclusion}
\[ x^{2007} + x^{-2007} = -1 - 1 = -2. \]
\[ (1 - 3x^2 + 2x^3) + (3x^2 - x^3) = 0. \] Step 2: {Solve for \( x \)}
\[ x^3 + 1 = 0. \] \[ x^3 = -1. \] \[ x = -\omega, -\omega^2, -1. \] Step 3: {Compute \( x^{2007} + x^{-2007} \)}
Since \( x^3 = -1 \), \[ x^{2007} = (-1)^{669} = -1. \] \[ x^{-2007} = -1. \] Step 4: {Conclusion}
\[ x^{2007} + x^{-2007} = -1 - 1 = -2. \]
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