A) a/c
B) - b/c
C) b/a
D) c/a
Correct Answer: B
Solution :
a, b, g are the roots of the equation \[{{x}^{3}}+a{{x}^{2}}+bx+c=0\] so \[\alpha +\beta +\gamma =-a\], \[\alpha \beta +\beta \gamma +\gamma \alpha =b\] and \[\alpha \beta \gamma =-\,c\] Now\[{{\alpha }^{-1}}+{{\beta }^{-1}}+{{\gamma }^{-1}}\] \[=\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }\] \[=\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }\] \[=-b/c\].
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