Multiple choice

If $\omega$ is a cube root of unity, then a root of the following equation is? $\begin{vmatrix} x+1 & \omega & \omega^2\ \omega & x+\omega^2 & 1\ \omega^2 & 1 & x+\omega \end{vmatrix}$.

  1. $x=1$
  2. $x=\omega$
  3. $x=\omega^2$
  4. $x=0$
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D Correct answer
Explanation

Substituting x=0 into the determinant results in a matrix where the sum of each row is 1+w+w^2 = 0. Since the sum of rows is zero, the determinant is zero, meaning x=0 is a root.

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