Question

If $f\left(x\right)=a+bx+{cx}^2$ and $\alpha ,\beta ,\gamma$ are the roots of the equation $x^3=1$, then $\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$ is equal to

• A. $f\left(\alpha \right)+f\left(\beta \right)+f\left(\gamma \right)$
• B. $f\left(\alpha \right)f\left(\beta \right)+f\left(\beta \right)f\left(\gamma \right)+f\left(\gamma \right)f\left(\alpha \right)$
• C. $f\left(\alpha \right)f\left(\beta \right)f\left(\gamma \right)$
• D. $-f\left(\alpha \right)f\left(\beta \right)f\left(\gamma \right)$

Answer 1

The correct option is D $-f\left(\alpha \right)f\left(\beta \right)f\left(\gamma \right)$

$f\left(x\right)=a+bx+{cx}^2$ and $\alpha ,\beta ,\gamma$ are the roots of the equation $x^3=1$
i.e $\alpha =1,\beta =\omega ,\gamma ={\omega }^2$ since $1,\omega ,{\omega }^2$ are cube roots of unity.
$\therefore \left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$
$=(a+b+c)\left|\begin{array}{ccc}1& 1& 1\\ b& c& a\\ c& a& b\end{array}\right|$
$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$=-(a+b+c)(a+b\omega +c{\omega }^2)(a+b{\omega }^2+{\omega }^4)$
$=-f\left(\alpha \right)f\left(\beta \right)f\left(\gamma \right)$
Hence, option D.