If the roots of $a{x}^4+4b{x}^3+6c{x}^2+4dx+e=0$  are equal in pairs then $b\left(3ac-2b^2\right)=$

$\alpha ,\alpha ,\beta ,\beta$

$a{x}^4+4b{x}^3+6c{x}^2+4dx+e=0$

$2\left(\alpha +\beta \right)=-4b$

$\alpha +\beta =\frac{-2b}{a},{\left(\alpha \beta \right)}^2=\frac{e}{a}\Rightarrow \alpha \beta =\sqrt{\frac{e}{a}}$

$\alpha ,\beta ,\alpha ,\beta$

$\alpha \beta +\alpha \beta +\alpha \beta +\beta \alpha +{\alpha }^2+{\beta }^2$

${\alpha }^2+{\beta }^2+4\alpha \beta =bca$

$\Rightarrow {\alpha }^2+{\beta }^2=\frac{6c}{a}-4\alpha \beta$

${\left(\alpha +\beta \right)}^2={\left(\frac{-2b}{a}\right)}^2$

${\left(\alpha +\beta \right)}^2=\frac{4b^2}{a^2}$

$\frac{4b^2}{a^2}=\frac{6c}{a}-4\alpha \beta +2\alpha \beta$

$\frac{6c}{a}-2\alpha \beta =\frac{4b^2}{a^2}$

$\frac{6c}{a}-\frac{4b^2}{a^2}=2\alpha \beta$

$\frac{3c}{a}-\frac{2b^2}{a^2}=\alpha \beta$

$\frac{3ac-b^2}{a^2}=\alpha \beta$

$3ac-2b^2=\left(\alpha \beta \right)a^2$

$\alpha \beta =\frac{3ac-2b^2}{a^2}$

${\alpha }^2\beta +{\alpha }^2\beta +{\beta }^2\alpha =\frac{-4d}{a}$

$2\left({\alpha }^2\beta +{\beta }^2\alpha \right)=\frac{-4d}{a}$

$\alpha \beta 2\left(\alpha +\beta \right)=\frac{-4d}{a}$

$2\alpha \beta \left(\alpha +\beta \right)=\frac{-4d}{a}$

$2\left(\frac{3ac-2b^2}{a^2}\right)\left(\frac{-2b}{a}\right)=\left(\frac{-4d}{2}\right)$

$\left(\frac{3ac-2b^2}{a^2}\right)\left(\frac{b}{a}\right)=\frac{d}{a}\times \frac{a}{a}$

$\left(\frac{3ac-2b^2}{a^2}\right)\left(\frac{b}{a}\right)=\frac{da}{a^2}$

$\left(3ac-2b\right)b=a^{2}d$