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

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

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given that $\alpha, \beta, \gamma$ are the roots of the equation $x^3+0x^2+4x+1=0$. 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: $\alpha+\beta = -\gamma$ $\beta+\gamma = -\alpha$ $\gamma+\alpha = -\beta$ Substituting these into the expression: $(\alpha+\beta)^{-1}+(\beta+\gamma)^{-1}+(\gamma+\alpha)^{-1} = \frac{1}{-\gamma} + \frac{1}{-\alpha} + \frac{1}{-\beta}$ $= -\left(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\right) = -\left(\frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma}\right)$ $= -\left(\frac{4}{-1}\right) = 4$

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

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