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

Using Vieta's formulas for $x^3+0x^2+4x+1=0$, $$\alpha+\beta+\gamma=0$$ $$\alpha\beta+\beta\gamma+\gamma\alpha=4$$ $$\alpha\beta\gamma=-1$$ Now, $$\alpha+\beta=-\gamma,\quad \beta+\gamma=-\alpha,\quad \gamma+\alpha=-\beta$$ So the required sum is $$-\left(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\right)$$ Further, $$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}=\frac{4}{-1}=-4$$ Hence the value is $$4$$ Checking the options, this matches $\text{(A)}$.