Given, $\alpha, \beta$ and $\gamma$ are the roots of
$\begin{aligned}
& x^3+4 x+1=0 . \\
& \begin{aligned}
& \therefore \alpha+\beta+\gamma=0, \alpha \beta+\beta \gamma+\gamma \alpha=4, \alpha \beta \gamma=-1 \\
& \text { Now, } \frac{\alpha^2}{\beta+\gamma}+\frac{\beta^2}{\gamma+\alpha}+\frac{\gamma^2}{\alpha+\beta}=\frac{\alpha^2}{-\alpha}+\frac{\beta^2}{-\beta}+\frac{\gamma^2}{-\gamma} \\
&=-(\alpha+\beta+\gamma)=0
\end{aligned} \\
& \frac{\alpha^2 \beta^2}{(\beta+\gamma)(\gamma+\alpha)}+\frac{\beta^2 \gamma^2}{(\gamma+\alpha)(\alpha+\beta)}+\frac{\gamma^2 \alpha^2}{(\beta+\gamma)(\alpha+\beta)}
\end{aligned}$
$\begin{aligned}=\alpha \beta+\beta \gamma+\gamma \alpha & =4 \\ \text { and } \quad \frac{\alpha^2 \beta^2 \gamma^2}{(\beta+\gamma)(\gamma+\alpha)(\alpha+\beta)} & =-\alpha \beta \gamma=1 \\ & (\because \alpha+\beta+\gamma=0)\end{aligned}$
$\therefore$ Required equation is
$x^3+4 x-1=0$