If \omega is a non-<strong>REAL</strong> cube root of unity, then what is a root of the following equation? \begin{vmatrix}x+1&\omega&\omega^{2}\\ \omega&x+\omega^{2}&1\\ \omega^{2}&1&x+\omega\end{vmatrix}=0

  1. A. x=0
  2. B. x=1
  3. C. x=\omega
  4. D. x=\omega^{2}

Correct Answer: A. x=0

Explanation

Applying C_1 \to C_1 + C_2 + C_3, the first column becomes x + 1 + \omega + \omega^2. Since 1 + \omega + \omega^2 = 0, the first column is simply x. Expanding along the first column shows x=0 is a root.

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