Question

If $\alpha ,\beta ,\gamma$ are the roots of the equation $x^3+px+q=0$ then the value of the determinant $\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{array}\right|$ is

• A. $q$
• B. $0$
• C. $p$
• D. $p^2-2q$

Answer 1

The correct option is B $0$

As $\alpha ,\beta ,\gamma$ are the roots of given equation

$⟹(\alpha +\beta +\gamma )=0$

Now,

$\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{array}\right|$

$R_1\to R_1+R_2+R_3$

$=\left|\begin{array}{ccc}\alpha +\beta +\gamma & \alpha +\beta +\gamma & \alpha +\beta +\gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{array}\right|$

$=(\alpha +\beta +\gamma )\left|\begin{array}{ccc}1& 1& 1\\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{array}\right|$

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

$⟹(\alpha +\beta +\gamma )\left(3\right(\alpha \beta +\alpha \gamma +\beta \gamma )-(\alpha +\beta +\gamma {)}^2)=0$ as $(\alpha +\beta +\gamma )=0$

Hence, option B is correct.