If $\alpha,\beta,\gamma$ are different from $1$ and are the roots of $ax^3+bx^2+cx+d=0$ and $(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)=\frac{25}{2}$,then $\sqrt[3]{\frac{-10(a+b+c+d)\Delta}{d}}$,where $\Delta=\begin{vmatrix}
\frac{\alpha}{1-\alpha} & \frac{\beta}{1-\beta} & \frac{\gamma}{1-\gamma} \\
\alpha & \beta & \gamma \\
\alpha^2 & \beta^2 & \gamma^2 \\
\end{vmatrix}$,is
I simplified $\Delta=(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)\alpha\beta\gamma(1-\alpha)(1-\beta)(1-\gamma)$
$=\frac{25}{2}\times \frac{-d}{a}(1+\frac{b}{a}+\frac{c}{a}+\frac{d}{a})$
Then $\sqrt[3]{\frac{-10(a+b+c+d)\Delta}{d}}=5\left(\frac{a+b+c+d}{a}\right)^{2/3}$
Then i could not solve it further.Its answer is an integer.Please help me.