Question

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

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

Answer 1

The correct option is D $x=0$

Let us substitute $x=0$, since it will eliminate one additional term ('x' term) in the expansion of the determinant.
If we substitute $x=0$, we get
$△=(2w^2-1)-w(2w-w^2)+w^2(w-w^4)$
$=2w^2-1-2w^2+w^3+w^3-w^6$
$=-1+2-1$ ...since $w^{3n}=1$
$=0$