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(C) Given that $a, b, c$ are the roots of the cubic equation $x^3+4x+1=0$. From Vieta's formulas,we have: $a+b+c = 0$ $ab+bc+ca = 4$ $abc = -1$ Since

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$4$

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(C) Given that $a, b, c$ are the roots of the cubic equation $x^3+4x+1=0$. From Vieta's formulas,we have: $a+b+c = 0$ $ab+bc+ca = 4$ $abc = -1$ Since $a+b+c = 0$,we can write: $a+b = -c$ $b+c = -a$ $c+a = -b$ Substituting these into the expression: $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} = \frac{1}{-c}+\frac{1}{-a}+\frac{1}{-b}$ $= -(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ $= -(\frac{bc+ac+ab}{abc})$ $= -(\frac{4}{-1}) = 4$

If $a, b$ and $c$ are the roots of $x^3+4x+1=0$,then $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=$

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