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:

• A. \( 1 \)
• B. \( -1 \)
• C. \( -2 \)
• D. \( 2 \)

Hint:

If \( x \) is a cube root of unity, then it satisfies \( x^3 = -1 \), which helps simplify large exponents.

Answer 1

The Correct Option is C

Step 1: {Expand the 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: {Calculate \( x^{2007} + x^{-2007} \)}
Given \( x^3 = -1 \),\[x^{2007} = (-1)^{669} = -1.\]\[x^{-2007} = -1.\]Step 4: {Final Result}
\[x^{2007} + x^{-2007} = -1 - 1 = -2.\]