A) 3
B) 4
C) 5
D) 6
Correct Answer: D
Solution :
Since\[\alpha ,\beta ,\gamma \]are the roots of the equation \[{{x}^{3}}+a{{x}^{2}}+b=0\] \[\therefore \] \[\left. \begin{align} & \alpha +\beta +\gamma =-a \\ & \alpha \beta +\beta \gamma +\gamma \alpha =0 \\ & \alpha \beta \gamma =-b \\ \end{align} \right]\] ...(i) Now, \[\left| \begin{matrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \\ \end{matrix} \right|\] \[=-(\alpha +\beta +\gamma )({{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}-\alpha \beta -\beta \gamma -\gamma \alpha )\] \[=-(\alpha +\beta +\gamma )[{{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}+2\alpha \beta +2\beta \gamma \] \[+2\gamma \alpha -3\alpha \beta -3\beta \gamma -3\gamma \alpha ]\] \[=-(\alpha +\beta +\gamma )[{{(\alpha +\beta +\gamma )}^{2}}\]\[-3(\alpha \beta +\beta \gamma +\gamma \alpha )]\] \[=-(-a)[{{a}^{2}}-0]\] [using Eq.(i)] \[={{a}^{3}}\]
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