question_answer

If\[\alpha ,\beta \]and\[\gamma \]are the roots of the equation \[{{x}^{3}}-8x+8=0,\]then\[\Sigma {{\alpha }^{2}}\]and\[\Sigma \frac{1}{\alpha \beta }\]are respectively

A)  0 and\[-16\]     

B)  16 and 18

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 ...(i) \[\therefore \]  \[{{(\alpha +\beta +\gamma )}^{2}}=0\] \[\Rightarrow \] \[{{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}+2(\alpha \beta +\beta \gamma +\gamma \alpha )=0\] \[\Rightarrow \]               \[\Sigma {{\alpha }^{2}}=-2(-8)\]                                   (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))