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

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Aicharya · Accepted Answer

$a, b, c$ are roots of $x^3+4 x+1=0$ $ \begin{aligned} \Rightarrow & x^3+0 \cdot x^2+4 x+1=0 \ 	ext { So, } & a+b+c=0 \Rightarrow a b c=-1 \end{aligned} $ $ 	ext { and } a b+b c+c a=4 $ Now, $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{-c}+\frac{1}{-a}+\frac{1}{-b}$ $ =-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}ight)=-\left(\frac{b c+a c+a b}{a b c}ight)=-\left(\frac{4}{-1}ight)=4 $

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

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