Question

If $\alpha$, $\beta$, $\gamma$ are the roots of $x^3+a{x}^2+b=0$, then the value of determinant $\Delta$ is , where
$\Delta =\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{array}\right|$.

• A. $-a^3$
• B. $a^3-3b$
• C. $a^2-3b$
• D. $a^3$

Answer 1

The correct option is D $a^3$

Value of determinant =$\Delta =\alpha (\beta .\gamma -{\alpha }^2)-\beta (\alpha .\gamma -{\beta }^2)+\gamma (\alpha .\beta -{\gamma }^2)$

$\to \Delta =3.\alpha .\beta .\gamma -({\alpha }^3+{\beta }^3+{\gamma }^3)$

$\to \alpha .\beta .\gamma =-b$

$\to \alpha +\beta +\gamma =-a$

$\to \alpha \beta +\beta \gamma +\gamma \alpha =0$

And ${\alpha }^3+{\beta }^3+{\gamma }^3-3\alpha \beta \gamma =(\alpha +\beta +\gamma )({\alpha }^2+{\beta }^2+{\gamma }^2-\alpha \beta -\beta \gamma -\gamma \alpha )$

Also ${\alpha }^2+{\beta }^2+{\gamma }^2=(\alpha +\beta +\gamma {)}^2-2(\alpha \beta +\beta \gamma +\gamma \alpha )$

Substituting we get ;

${\alpha }^3+{\beta }^3+{\gamma }^3=3(-b)+(-a)\left(a^2\right)$

$\therefore \Delta =a^3+3b-3b$

$\Rightarrow \Delta =a^3$