Question

If $\omega$ is imaginary cube root of unity, then a root of the equation $\left|\begin{array}{ccc}x+1& \omega & {\omega }^2\\ \omega & x+{\omega }^2& 1\\ {\omega }^2& 1& x+\omega \end{array}\right|=0$ is

• A. $x=1$
• B. $x=\omega$
• C. $x={\omega }^2$
• D. $x=0$

Answer 1

Answer: (D)

Put $x=0(x=0)\Rightarrow \left|\begin{array}{ccc}1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \end{array}\right|=0$