opepayflarpws

2022-06-20

If $\alpha ,\beta ,\gamma$ are roots of the cubic equation $2{x}^{3}+3{x}^{2}-x-1=0$

Aaron Everett

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

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