Question

If $\alpha ,\beta ,\gamma$ are the roots of $a{x}^3+b{x}^2+cx+d=0$ and $\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{array}\right|=0$,$\alpha \ne \beta \ne \gamma$, then find the equation whose roots are $\alpha +\beta -\gamma ,\gamma +\alpha -\beta ,\beta +\gamma -\alpha$

• A. $a{y}^3-2b{y}^2-4cy-8d=0$
• B. $a{x}^3-2b{x}^2+4cx+8d=0$
• C. $a{x}^3-2b{x}^2+4cx-8d=0$
• D. $a{y}^3-2b{y}^2+4cy-8d=0$

Answer 1

The correct option is D $a{y}^3-2b{y}^2+4cy-8d=0$

Given $\alpha ,\beta ,\gamma$ are the roots of $a{x}^3+b{x}^2+cx+d=0$ and
$\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{array}\right|=0$,$\alpha \ne \beta \ne \gamma$
$\Rightarrow (\alpha +\beta +\gamma )\left|\begin{array}{ccc}1& 1& 1\\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{array}\right|=0$
$\Rightarrow (\alpha +\beta +\gamma )(\alpha \beta +\beta \gamma +\gamma \alpha -{\alpha }^2-{\beta }^2-{\gamma }^2)=0$
$\Rightarrow \frac{-1}{2}(\alpha +\beta +\gamma )\left(\right(\alpha -\beta {)}^2+(\beta -\gamma {)}^2+(\gamma -\alpha {)}^2)=0$
$\therefore \alpha +\beta +\gamma =0$ as $\alpha \ne \beta \ne \gamma$
$y=\alpha +\beta -\gamma =-2\gamma$
$\Rightarrow \gamma =\frac{-y}{2}$
$\therefore$ Equation whose roots are $\alpha +\beta -\gamma ,\gamma +\alpha -\beta ,\beta +\gamma -\alpha$ is
$a{\left(\frac{-y}{2}\right)}^3+b{\left(\frac{-y}{2}\right)}^2+c\left(\frac{-y}{2}\right)+d=0$
$a{y}^3-2b{y}^2+4cy-8d=0$
Hence, options D.