If $\alpha ,\beta ,\gamma$ are the roots of x³ – 6x – 4 = 0 then the equation whose roots are $\mathrm{\beta \gamma }+\frac{1}{\mathrm{\alpha }},\mathrm{\gamma \alpha }+\frac{1}{\mathrm{\beta }},\mathrm{\alpha \beta }+\frac{1}{\mathrm{\gamma }}$ is

x³ – 6x – 4 = 0

$$\begin{gathered} \alpha ,\beta ,\gamma are roots then, \\ \alpha +\beta +\gamma =0 \\ \alpha \beta +\beta \gamma +\gamma \alpha =-6 \\ \alpha \beta \gamma =4 \end{gathered}$$

$\frac{\mathrm{\alpha \beta \gamma }+1}{\mathrm{\alpha }},\frac{\mathrm{\alpha \beta \gamma }+1}{\mathrm{\beta }},\frac{\mathrm{\alpha \beta \gamma }+1}{\mathrm{\gamma }}$ are roots
$$\begin{gathered} \text{ i.e., }\frac{5}{\mathrm{\alpha }},\frac{5}{\mathrm{\beta }},\frac{5}{\mathrm{\gamma }}\text{ roots } \\ \left(i\right) \\ \frac{5}{\gamma }+\frac{5}{\beta }+\frac{5}{\alpha }=\frac{-15}{2} \\ \left(ii\right) \\ \frac{5}{\gamma }·\frac{5}{\beta }+\frac{5}{\beta }·\frac{5}{\alpha }+\frac{5}{\gamma }·\frac{5}{\alpha }=0 \\ \left(iii\right) \\ \frac{5}{\gamma }·\frac{5}{\beta }·\frac{5}{\alpha }=\frac{125}{4} \\ then the equation is \\ {x}^3-x^2\left(\frac{{\displaystyle 5}}{{\displaystyle \gamma }}+\frac{{\displaystyle 5}}{{\displaystyle \beta }}+\frac{{\displaystyle 5}}{{\displaystyle \alpha }}\right)+x\left(\frac{{\displaystyle 5}}{{\displaystyle \gamma }}·\frac{{\displaystyle 5}}{{\displaystyle \beta }}+\frac{{\displaystyle 5}}{{\displaystyle \beta }}·\frac{{\displaystyle 5}}{{\displaystyle \alpha }}+\frac{{\displaystyle 5}}{{\displaystyle \gamma }}·\frac{{\displaystyle 5}}{{\displaystyle \alpha }}\right)-\left(\frac{{\displaystyle 5}}{{\displaystyle \gamma }}·\frac{{\displaystyle 5}}{{\displaystyle \beta }}·\frac{{\displaystyle 5}}{{\displaystyle \alpha }}\right)=0 \\ {x}^3+\left(\frac{-15}{2}\right)x^2+\left(0\right)x+\left(\frac{125}{4}\right)=0 \\ 4x^3-30x^2+125=0 \end{gathered}$$