Question

Let $f\left(x\right)=a+bx+c{x}^2$ and $\omega \ne 1$ be a cube root of unity, and
$\Delta =\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$
then factors of $\Delta$ are

• A. f(1)
• B. $f\left(\omega \right)$
• C. $f\left({\omega }^2\right)$
• D. none of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A f(1)
B $f\left(\omega \right)$
C $f\left({\omega }^2\right)$

$\Delta =\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$

Applying $C_1\to C_1+C_2+C_3$

$\Delta =\left|\begin{array}{ccc}a+b+c& b& c\\ b+c+a& c& a\\ c+a+b& a& b\end{array}\right|=(a+b+c)\left|\begin{array}{ccc}1& b& c\\ 1& c& a\\ 1& a& b\end{array}\right|$

Applying $R_2\to R_2-R_1,R_3\to R_3-R_1$

$\Delta =(a+b+c)\left|\begin{array}{ccc}1& b& c\\ 0& c-b& a-c\\ 0& a-b& b-c\end{array}\right|=(a+b+c)(-{(c-b)}^2-(a-b)(a-c))$

$=(a+b+c)(-a^2-b^2-c^2+ab+ac+bc)$

$=(a+b+c)(a{w}^2+b\omega +c)(-a+b{\omega }^2+c)$

Now

$f\left(1\right)=a+b+c$

$f\left(\omega \right)=a{w}^2+b\omega +c$

$f\left({\omega }^2\right)=a{\omega }^4+b{\omega }^2+c=-a+b{\omega }^2+c$