Question

If $f\left(x\right)=a+bx+c{x}^2$ 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. $f\left(\alpha \right)f\left(\beta \right)+f\left(\beta \right)f\left(\gamma \right)+f\left(\gamma \right)f\left(\alpha \right)$
• B. $f\left(\alpha \right)+f\left(\beta \right)+f\left(\gamma \right)$
• 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 $-f\left(\alpha \right)f\left(\beta \right)f\left(\gamma \right)$

$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=-(a^3+b^3+c^3-3abc)$

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

(Where $\omega$ is cube root of unity)

$=-f\left(\alpha \right)f\left(\beta \right)f\left(\gamma \right)[\because \alpha =1;\beta =\omega ;\gamma ={\omega }^2]$