If the equation ${\mathrm{x}}^4+{\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+\mathrm{d}=0$ has three equal roots , then that root is 

${\mathrm{x}}^4+{\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+\mathrm{d}=0$  has three equal roots then

$S_1=3\alpha +\beta =-a;S_2=3\alpha (\alpha +\beta )=b;S_3={\alpha }^2(\alpha +3\beta )=-c;S_4={\alpha }^3\beta$

$\begin{array}{r}6c-ab=\alpha \left(3{\alpha }^2-6\alpha \beta +3{\beta }^2\right)\\ 3a^2-8b=3{\alpha }^2-6\alpha \beta +3{\beta }^2\end{array}$

$\therefore \frac{6c-ab}{3a^2-8b}=\frac{\alpha \left(3{\alpha }^2-6a\beta +3{\beta }^2\right)}{3{\alpha }^2-6\alpha \beta +3{\beta }^2}=\alpha$