If alpha, beta, gamma are the roots of the eq | vedclass

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(C) Given the cubic equation $x^3+a x^2+b x+c=0$ with roots $\alpha, \beta, \gamma$. By Vieta's formulas: $\alpha+\beta+\gamma = -a$ $\alpha \beta+\be

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$-\frac{b}{c}$

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(C) Given the cubic equation $x^3+a x^2+b x+c=0$ with roots $\alpha, \beta, \gamma$. By Vieta's formulas: $\alpha+\beta+\gamma = -a$ $\alpha \beta+\beta \gamma+\gamma \alpha = b$ $\alpha \beta \gamma = -c$ We need to find $\alpha^{-1}+\beta^{-1}+\gamma^{-1} = \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$. Combining the terms: $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} = \frac{\alpha \beta+\beta \gamma+\gamma \alpha}{\alpha \beta \gamma}$. Substituting the values from Vieta's formulas: $\frac{b}{-c} = -\frac{b}{c}$.

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

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