If $\mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma }$ are the roots of ${\mathrm{x}}^3+4\mathrm{x}+1=0,$ then the equation whose roots are $\frac{{\mathrm{\alpha }}^2}{\mathrm{\beta }+\mathrm{\gamma }},\frac{{\mathrm{\beta }}^2}{\mathrm{\gamma }+\mathrm{\alpha }},\frac{{\mathrm{\gamma }}^2}{\mathrm{\alpha }+\mathrm{\beta }}$ is

$\begin{array}{l}If\alpha ,\beta ,\gamma arerootsof{x}^3+4x+1=0\\ Then\alpha +\beta +\gamma =0,\alpha \beta +\beta \gamma +\gamma \alpha =4,\alpha \beta \gamma =-1\\ Let\frac{{\alpha }^2}{\beta +\gamma }=y\Rightarrow \frac{{\alpha }^2}{-\alpha }=y\Rightarrow \alpha =-y\\ \because \alpha =-yisrootof{x}^3+4x+1=0\\ \Rightarrow {(-y)}^3+4(-y)+1=0\\ \Rightarrow -y^3-4y+1=0\\ \Rightarrow y^3+4y-1=0\\ \mathrm{Re}quiredEquationis{x}^3+4x-1=0\end{array}$