Question

If ω is a cube root of unity, then $\left|\begin{array}{ccc}x+1 & \omega & a^2 \\ \omega & x+\omega^2 & 1 \\ \omega^2 & 1 & x+\omega\end{array}\right|$ equal to

(A) $x^2+1$

(B) $x^3+()$

(C) $x^3+\omega^2$

(D) $x^3$

Answer 1

Applying $R_1 \rightarrow R_1+R_2+R_3$ and taking common from $R_1$, we get

$\begin{array}{l}\left(x+1+\omega+\omega^2\right)\left|\begin{array}{ccc}1 & 1 & 1 \\\omega & x+\omega^2 & 1 \\\omega^2 & 1 & x+\omega\end{array}\right| \\=x\left|\begin{array}{ccc}1 & 0 & 0 \\\omega & x+\omega^2-\omega & 1-\omega \\\omega^2 & 1-\omega^2 & x+\omega-\omega^2\end{array}\right| \\=x\left[\left(x+\omega^2-\omega\right)\left(x+\omega-\omega^2\right)\right. \\=x^3\end{array}$

So, The correct option is (D).