A) 0 and-16
B) 16 and 8
C) -16 and 0
D) 16 and 0
Correct Answer: D
Solution :
Since \[\alpha ,\beta \] and \[\gamma \]are the roots of the equation \[{{x}^{3}}-8x+8=0,\] then \[\alpha +\beta +\gamma =0,\] \[\alpha \beta +\beta \gamma +\gamma \alpha =-8,\] \[\alpha \beta \gamma =-8\] ...(i) \[\therefore \] s\[{{(\alpha +\beta +\gamma )}^{2}}=0\] \[\Rightarrow \] \[{{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}+2(\alpha \beta +\beta \gamma +\gamma \alpha )=0\] \[\Rightarrow \] \[\Sigma {{\alpha }^{2}}=-2(-8)\] (From (i)) \[=16\] and \[\frac{1}{\alpha \beta }+\frac{1}{\beta \gamma }+\frac{1}{\gamma \alpha }=\frac{\gamma +\alpha +\beta }{\alpha \beta \gamma }\] \[\Rightarrow \] \[\frac{1}{\Sigma \alpha \beta }=\frac{0}{-8}=0\] (From (i))
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