If α , β , γ are roots of the cubic equation 2 x 3...

If
$2x^3+3x^2-x-1=0$
has roots $\alpha ,\beta ,\gamma$, then substituting $x\mapsto \frac{1}{x}$ (and multiplying by $-x^3$ to clear denominators)
$\begin{array}{rl}& -x^3(\frac{2}{x^3}+\frac{3}{x^2}-\frac{1}{x}-1)\\ & =x^3+x^2-3x-2=0\end{array}$
has roots $\frac{1}{\alpha },\frac{1}{\beta },\frac{1}{\gamma }$. Then substituting $x\mapsto -\frac{2}{3}x$ (and multiplying by $-\frac{27}{8}$ to clear denominators)
$\begin{array}{rl}& -\frac{27}{8}({(-\frac{2}{3}x)}^3+{(-\frac{2}{3}x)}^2-3(-\frac{2}{3}x)-2)\\ & x^3-\frac{3}{2}{x}^2-\frac{27}{4}x+\frac{27}{4}=0\end{array}$
has roots $\frac{-3/2}{\alpha }=\frac{\alpha +\beta +\gamma }{\alpha },\frac{-3/2}{\beta }=\frac{\alpha +\beta +\gamma }{\beta },\frac{-3/2}{\gamma }=\frac{\alpha +\beta +\gamma }{\gamma }$. Next, substituting $x\mapsto x+1$
$\begin{array}{rl}& (x+1{)}^3-\frac{3}{2}(x+1{)}^2-\frac{27}{4}(x+1)+\frac{27}{4}\\ & =x^3+\frac{3}{2}{x}^2-\frac{27}{4}x-\frac{1}{2}=0\end{array}$
has roots $\frac{\beta +\gamma }{\alpha },\frac{\alpha +\gamma }{\beta },\frac{\alpha +\beta }{\gamma }$. Finally, substituting $x\mapsto \frac{1}{x}$ (and multiplying by $-4x^3$ to clear denominators)
$\begin{array}{rl}& -4x^3(\frac{1}{x^3}+\frac{3}{2}\frac{1}{x^2}-\frac{27}{4}\frac{1}{x}-\frac{1}{2})\\ & =2x^3+27x^2-6x-4=0\end{array}$
has roots $\frac{\alpha }{\beta +\gamma },\frac{\beta }{\alpha +\gamma },\frac{\gamma }{\alpha +\beta }$