A) \[-\frac{1}{2}\]
B) \[-1\]
C) 0
D) \[\frac{1}{2}\]
Correct Answer: B
Solution :
Given cubic equation is, \[{{x}^{3}}-2x+1=0,(\alpha ,\beta ,\gamma )\]are roots of this equation. Then sum of roots \[\sum{\alpha =0}\] \[\Rightarrow \]\[\alpha +\beta +\gamma =0\] \[\sum{\alpha \beta }=-2,\alpha \beta \gamma =-1\] Now, we have \[\sum{\frac{1}{^{\alpha +\beta -\gamma }}}\sum{\frac{1}{-\gamma -\gamma }}=-\frac{1}{2}\sum{\frac{1}{\gamma }}\] \[=-\frac{1}{2}\left( \frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma } \right)\] \[=-\frac{1}{2}\left( \frac{\alpha \beta +\beta \gamma +\alpha \gamma }{\alpha \beta \gamma } \right)\] \[=-\frac{1}{2}.\frac{\sum{\alpha \beta }}{\alpha \beta \gamma }=-\frac{1}{2}.\frac{(-2)}{(-1)}=-1\]
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