A) 80
B) 84
C) 90
D) - 84
Correct Answer: B
Solution :
Since, \[\alpha ,\beta ,\gamma \]are the roots of the equation \[{{x}^{3}}-6{{x}^{2}}+11x+6=0\] \[\therefore \] \[\alpha +\beta +\gamma =6\] \[\alpha \beta +\beta \gamma +\gamma \alpha =11\] and \[\alpha \beta \gamma =-6\] Now, \[\sum {{\alpha }^{2}}\beta +\sum \alpha {{\beta }^{2}}\] \[=\sum {{\alpha }^{2}}\beta +\sum \alpha {{\beta }^{2}}\] \[={{\alpha }^{2}}\beta +{{\beta }^{2}}\gamma +{{\gamma }^{2}}\alpha +\alpha {{\beta }^{2}}+\beta {{\gamma }^{2}}+\gamma {{\alpha }^{2}}\] \[=\alpha \beta (\alpha +\beta )+\beta \gamma (\beta +\gamma )+\gamma \alpha (\gamma +\alpha )\] \[=\alpha \beta (6-\gamma )+\beta \gamma (6-\alpha )+\gamma \alpha (6-\beta )\] \[=6(\alpha \beta +\beta \gamma +\gamma \alpha )-3\alpha \beta \gamma \] \[=6(11)+3(6)\] \[=66+18=84\]
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