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

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

Vedclass · Accepted Answer

$-\frac{b}{c}$

Vedclass · Answer

(C) Given the cubic equation $x^3+ax^2+bx+c=0$ 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 $\alpha^{-1}+\beta^{-1}+\gamma^{-1} = \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$. Taking the common denominator: $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} = \frac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}$. Substituting the values from Vieta's formulas: $\frac{b}{-c} = -\frac{b}{c}$.

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

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