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(B) Given that $\alpha, \beta, \gamma$ are the roots of the cubic equation $x^3+a x^2+b x+c=0$. By Vieta's formulas,the sum of the roots taken two at

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$-\frac{b}{c}$

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(B) Given that $\alpha, \beta, \gamma$ are the roots of the cubic equation $x^3+a x^2+b x+c=0$. By Vieta's formulas,the sum of the roots taken two at a time is $\alpha \beta + \beta \gamma + \gamma \alpha = \frac{b}{1} = b$,and the product of the roots is $\alpha \beta \gamma = -\frac{c}{1} = -c$. We need to find the value of $\alpha^{-1}+\beta^{-1}+\gamma^{-1}$. $\alpha^{-1}+\beta^{-1}+\gamma^{-1} = \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta \gamma + \alpha \gamma + \alpha \beta}{\alpha \beta \gamma}$. Substituting the values,we get $\frac{b}{-c} = -\frac{b}{c}$.

If $\alpha, \beta, \gamma$ are roots of the equation $x^3+a x^2+b x+c=0$,then $\alpha^{-1}+\beta^{-1}+\gamma^{-1} = $

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