relation between roots and coefficient in a cubic polynomial If $\alpha,\beta,\gamma$ are roots of the cubic equation
$$2x^{3}+3x^2-x-1=0$$ then I want to find the equation whose roots are $\frac{\alpha}{\beta+\gamma}, \frac{\beta}{\gamma+\alpha}, \frac{\gamma}{\alpha+\beta}$.
I have $\alpha+\beta+\gamma$= - $\frac{3}{2}$.
Therefore roots become $\frac{\alpha}{-\frac{3}{2} -\alpha}$,$\frac{\beta}{-\frac{3}{2} -\beta}$, $\frac{\gamma}{-\frac{3}{2} -\gamma}$.
To find the equation we have to find their sums and products, but it looks like complicated.
Is their any easy trick ?

If
$$
2x^3+3x^2-x-1=0
$$
has roots $\alpha,\beta,\gamma$, then substituting $x\mapsto\frac1x$ (and multiplying by $-x^3$ to clear denominators)
$$
\begin{align}
&-x^3\left(\frac2{x^3}+\frac3{x^2}-\frac1x-1\right)\\
&=x^3+x^2-3x-2=0
\end{align}
$$
has roots $\frac1\alpha,\frac1\beta,\frac1\gamma$. Then substituting $x\mapsto-\frac23x$ (and multiplying by $-\frac{27}8$ to clear denominators)
$$
\begin{align}
&-\frac{27}8\left(\left(-\frac23x\right)^3+\left(-\frac23x\right)^2-3\left(-\frac23x\right)-2\right)\\
&x^3-\frac32x^2-\frac{27}4x+\frac{27}4=0
\end{align}
$$
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 $x\mapsto x+1$
$$
\begin{align}
&(x+1)^3-\frac32(x+1)^2-\frac{27}4(x+1)+\frac{27}4\\
&=x^3+\frac32x^2-\frac{27}4x-\frac12=0
\end{align}
$$
has roots $\frac{\beta+\gamma}\alpha,\frac{\alpha+\gamma}\beta,\frac{\alpha+\beta}\gamma$. Finally, substituting $x\mapsto\frac1x$ (and multiplying by $-4x^3$ to clear denominators)
$$
\begin{align}
&-4x^3\left(\frac1{x^3}+\frac32\frac1{x^2}-\frac{27}4\frac1x-\frac12\right)\\
&=2x^3+27x^2-6x-4=0
\end{align}
$$
has roots $\frac\alpha{\beta+\gamma},\frac{\beta}{\alpha+\gamma},\frac{\gamma}{\alpha+\beta}$.
