Let the roots of $x^3+b x+c=0$ is
$$
\begin{aligned}
& \cos \alpha+i \sin \alpha, \cos \beta+i \sin \beta \text { and } \cos \gamma+i \sin \gamma \\
& \begin{aligned}
& \text { Sum of roots }=\cos \alpha+i \sin \alpha+\cos \beta \\
&+i \sin \beta+\cos \gamma+i \sin \gamma \\
& \Rightarrow 0=(\cos \alpha+\cos \beta+\cos \gamma+i(\sin \alpha+\sin \beta+\sin \gamma) \\
& \text { Hence, } \cos \alpha+\cos \beta+\cos \gamma=0
\end{aligned} \\
& \text { and } \sin \alpha+\sin \beta+\sin \gamma=0 \\
& \Rightarrow \quad 2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)=-\cos \gamma \\
& \text { and } 2 \sin \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)=-\sin \gamma
\end{aligned}
$$
(i) divided by (ii), we get
$$
\begin{gathered}
\cot \left(\frac{\alpha+\beta}{2}\right)=\cot \gamma \\
\alpha+\beta=2 \gamma
\end{gathered}
$$


Similarly, $\beta+\gamma=2 \alpha$ and $\gamma+\alpha=2 \beta$ and $\alpha=\beta=\gamma$
$$
\begin{array}{rlrl}
& \therefore & & \cos \alpha+\cos \beta+\cos \gamma=0 \\
\Rightarrow & & 3 \cos \alpha=0 \\
\Rightarrow & \cos \alpha=\cos \beta=\cos \gamma=0
\end{array}
$$
Now, $b=\cos (\alpha+\beta)+\cos (\beta+\gamma)+\cos (\gamma+\alpha)$
$$
+1[\sin (\alpha+\beta)+\sin (\beta+\gamma)+\sin (\gamma+\alpha)]
$$
$\therefore$ Real part of $b$
$$
\begin{aligned}
& =\cos (\alpha+\beta)+\cos (\beta+\gamma)+\cos (\alpha+\gamma) \\
& =\cos 2 \gamma+\cos 2 \alpha+\cos 2 \beta \\
& =2 \cos ^2 \gamma-1+2 \cos ^2 \alpha-1+2 \cos ^2 \beta-1 \\
& =0-1+0-1+0-1=-3
\end{aligned}
$$
Hence, no option is matched.