If $\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma }$ are the roots of x3 + x – 2 = 0 then the equation whose roots are  $\frac{\mathrm{\beta }}{\mathrm{\gamma }}+\frac{\mathrm{\gamma }}{\mathrm{\beta }},\frac{\mathrm{\gamma }}{\mathrm{\alpha }}+\frac{\mathrm{\alpha }}{\mathrm{\gamma }},\frac{\mathrm{\alpha }}{\mathrm{\beta }}+\frac{\mathrm{\beta }}{\mathrm{\alpha }}$ is

$$\begin{gathered} \begin{array}{l}\frac{\mathrm{\beta }}{\mathrm{\gamma }}+\frac{\mathrm{\gamma }}{\mathrm{\beta }}=\frac{{\mathrm{\beta }}^2+{\mathrm{\gamma }}^2}{\mathrm{\gamma \beta }}=\frac{(\mathrm{\beta }+\mathrm{\gamma }{)}^2-2\mathrm{\beta \gamma }}{\mathrm{\gamma \beta }}\\ =\frac{{\mathrm{\alpha }}^2-2\mathrm{\beta \gamma }}{\mathrm{\gamma \beta }}=\frac{{\mathrm{\alpha }}^3-2\mathrm{\alpha \beta \gamma }}{\mathrm{\alpha \beta \gamma }}=\frac{{\mathrm{\alpha }}^3-4}{2}\\ {\mathrm{\alpha }}^3=2-\mathrm{\alpha }\Rightarrow \frac{2-\mathrm{\alpha }-4}{2}=\frac{-\mathrm{\alpha }-2}{2}=\frac{-\mathrm{\alpha }}{2}-1=\mathrm{x} \\ \Rightarrow \mathrm{\alpha }=-2(\mathrm{x}+1)\\ \therefore \mathrm{take} \mathrm{f}(-2(\mathrm{x}+1\left)\right)=0 then,\\ \Rightarrow 4x^3+12x^2+13x+6=0\end{array} \end{gathered}$$