If $\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma }$ are the roots of x³–3x+1=0 then the equation whose roots are $\mathrm{\alpha }-\frac{1}{\mathrm{\beta \gamma }},\mathrm{\beta }-\frac{1}{\mathrm{\gamma \alpha }},\mathrm{\gamma }-\frac{1}{\mathrm{\alpha \beta }}$ is

$\begin{array}{r}\mathrm{\alpha }-\frac{1}{\mathrm{\beta \gamma }}=\mathrm{\alpha }\left(1-\frac{1}{\mathrm{\alpha \beta \gamma }}\right)=\mathrm{\alpha }\left(1-\frac{1}{-1}\right)=2\mathrm{\alpha }\Rightarrow \end{array}$

$\begin{array}{r}If \alpha , \beta , \gamma are the roots of f\left(x\right)=x^3–3x+1=0 \\ then the equation whose roots are\mathrm{\alpha }-\frac{1}{\mathrm{\beta \gamma }} ,\mathrm{\beta }-\frac{1}{\mathrm{\gamma \alpha }},\mathrm{\gamma }-\frac{1}{\mathrm{\alpha \beta }} (i.e)\\ 2\mathrm{\alpha },2\mathrm{\beta },2\mathrm{\gamma }\text{ are roots of }\mathrm{f}\left(\frac{\mathrm{x}}{2}\right)=0\\ f\left(\frac{x}{2}\right)={\left(\frac{x}{2}\right)}^3-3\left(\frac{x}{2}\right)+1=0\Rightarrow x^3-12x+8=0\end{array}$