A) \[p\]
B) \[q\]
C) \[{{p}^{2}}-2q\]
D) \[0\]
Correct Answer: D
Solution :
Since,\[\alpha ,\beta ,\gamma \]are the roots of \[{{x}^{3}}+px+q=0\] \[\therefore \] \[\alpha +\beta +\gamma =0\] \[\therefore \] \[\left| \begin{matrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \\ \end{matrix} \right|\] \[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}},\]we get \[=\left| \begin{matrix} \alpha +\beta +\gamma & \alpha +\beta +\gamma & \alpha +\beta +\gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \\ \end{matrix} \right|=0\]
You need to login to perform this action.
You will be redirected in
3 sec
