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

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(C) Given the cubic equation $x^3 + 0x^2 + 4x + 1 = 0$,the roots are $\alpha, \beta, \gamma$.From Vieta's formulas,we have:$\alpha + \beta + \gamma =

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$4$

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(C) Given the cubic equation $x^3 + 0x^2 + 4x + 1 = 0$,the roots are $\alpha, \beta, \gamma$.From Vieta's formulas,we have:$\alpha + \beta + \gamma = 0$$\alpha\beta + \beta\gamma + \gamma\alpha = 4$$\alpha\beta\gamma = -1$Since $\alpha + \beta + \gamma = 0$,we can write $\alpha + \beta = -\gamma$,$\beta + \gamma = -\alpha$,and $\gamma + \alpha = -\beta$.Substituting these into the expression:$(\alpha + \beta)^{-1} + (\beta + \gamma)^{-1} + (\gamma + \alpha)^{-1} = \frac{1}{-\gamma} + \frac{1}{-\alpha} + \frac{1}{-\beta} = -(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma})$$= -(\frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma})$$= -(\frac{4}{-1}) = 4$.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3 + 4x + 1 = 0$,then $(\alpha + \beta)^{-1} + (\beta + \gamma)^{-1} + (\gamma + \alpha)^{-1} = $

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