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

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(B) Given that $\alpha, \beta, \gamma$ are the roots of the cubic equation $x^3 + ax^2 + bx + c = 0$.According to Vieta's formulas for a cubic equatio

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$-b/c$

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(B) Given that $\alpha, \beta, \gamma$ are the roots of the cubic equation $x^3 + ax^2 + bx + c = 0$.According to Vieta's formulas for a cubic equation $Ax^3 + Bx^2 + Cx + D = 0$:Sum of roots: $\alpha + \beta + \gamma = -a$Sum of roots taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = b$Product of roots: $\alpha\beta\gamma = -c$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}$Substituting the values from Vieta's formulas:$= \frac{b}{-c} = -\frac{b}{c}$.

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

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