Question

If $\omega$ is cube root of unity, then
$\Delta =\left|\begin{array}{c}x+1\\ \omega \\ {\omega }^2\end{array}\begin{array}{c}\omega \\ x+{\omega }^2\\ 1\end{array}\begin{array}{c}{\omega }^2\\ 1\\ x+\omega \end{array}\right|=$

• A. $x^3+1$
• B. $x^3+\omega$
• C. $x^3+{\omega }^2$
• D. $x^3$

Answer 1

Answer: (D)

$x^3$

$△=\left|\begin{array}{ccc}x+1& \omega & {\omega }^2\\ \omega & x+{\omega }^2& 1\\ {\omega }^2& 1& x+\omega \end{array}\right|$

$1+\omega +{\omega }^2=0{\omega }^3=1$

$R_1\to R_1+R_2+R_3$

$△=\left|\begin{array}{ccc}1+x+\omega +{\omega }^2& 1+\omega +{\omega }^2+x& 1+\omega +{\omega }^2+x\\ \omega & x+{\omega }^2& 1\\ {\omega }^2& 1& x+\omega \end{array}\right|$

$△=x\left|\begin{array}{ccc}1& 1& 1\\ \omega & x+{\omega }^2& 1\\ {\omega }^2& 1& x+\omega \end{array}\right|$

$△=x[(x+{\omega }^2)(x+\omega )-1(x\omega +{\omega }^2-x{\omega }^2-\omega )(\omega -x{\omega }^2-\omega )]$

$△=x[x^2+x(\omega +{\omega }^2)+1-1\left(x\omega \right)+1(-x{\omega }^2)]$

$\Rightarrow △=x^3$

Option D