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. $f\left(\alpha \right)+f\left(\beta \right)+f\left(\gamma \right)$
• 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 $-f\left(\alpha \right)f\left(\beta \right)f\left(\gamma \right)$

Given,

$f\left(x\right)=a+bx+c{x}^2$

and $\alpha ,\beta ,\gamma$ are roots $x^3=1$

which means $\alpha ,\beta ,\gamma$ are $1,w,w^2$

and we know , $w^3=1$ and $1+w+w^2=0$

so, $f\left(\alpha \right)=f\left(1\right)=a+b+c$ ,$f\left(\beta \right)=f\left(w\right)=a+bw+c{w}^2$ and $f\left(\gamma \right)=f\left(w^2\right)=a+b{w}^2+cw$

now,

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

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

$=abc-a^3+abc-b^3+abc-c^3$

$=3abc-a^3-b^3-c^3$

$=-(a^3+b^3+c^3-3abc)$

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

$=-(a+b+c)(a^2+b^2+c^2+(-1)ab+(-1)bc+(-1\left)ca\right)$

$=-(a+b+c)(a^2+b^2+c^2+(w+w^2)ab+(w+w^2)bc+(w+w^2\left)ca\right)$

$=-(a+b+c)(a+bw+c{w}^2)(a+b{w}^2+cw)$

$=-f\left(1\right)f\left(w\right)f\left(w^2\right)$

$=-f\left(\alpha \right)f\left(\beta \right)f\left(\gamma \right)$