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(C) Given that $\alpha, \beta, \gamma$ are the roots of the cubic equation $x^3+0x^2+4x+1=0$. By Vieta's formulas,the sum of the roots is $\alpha+\bet

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$4$

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(C) Given that $\alpha, \beta, \gamma$ are the roots of the cubic equation $x^3+0x^2+4x+1=0$. By Vieta's formulas,the sum of the roots is $\alpha+\beta+\gamma = 0$. Therefore,$\alpha+\beta = -\gamma$,$\beta+\gamma = -\alpha$,and $\gamma+\alpha = -\beta$. The expression becomes: $(\alpha+\beta)^{-1}+(\beta+\gamma)^{-1}+(\gamma+\alpha)^{-1} = \frac{1}{-\gamma} + \frac{1}{-\alpha} + \frac{1}{-\beta} = -(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma})$. This simplifies to: $-(\frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma})$. From the equation $x^3+0x^2+4x+1=0$,we have $\alpha\beta + \beta\gamma + \gamma\alpha = 4$ and $\alpha\beta\gamma = -1$. Substituting these values: $-(\frac{4}{-1}) = -(-4) = 4$.

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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