If the equations ${\mathrm{ax}}^3+3{\mathrm{bx}}^2+3\mathrm{cx}+\mathrm{d}=0\text{ and }{\mathrm{ax}}^2+2\mathrm{bx}+\mathrm{c}=0$ have a common root, then

The given equations are $f\left(x\right)={\mathrm{ax}}^3+3{\mathrm{bx}}^2+3\mathrm{cx}+\mathrm{d}=0\text{ and g}\left(x\right)\text{=}{\mathrm{ax}}^2+2\mathrm{bx}+\mathrm{c}=0$

Given that $x=\alpha$ is a common root to the above two equations $f\left(x\right)=0$ and $g\left(x\right)=0$

The root $x=\alpha$ is a root of the function $f\left(x\right)-g\left(x\right)=0$

Multiplying the function $g\left(x\right)$ by $x$ subtract from $f\left(x\right)$

$a{x}^3+3b{x}^2+3cx+d-a{x}^3-2b{x}^2-cx=0\Rightarrow b{x}^2+2cx+d=0$ -----(1)

Multiply the equation (1) with a and then subtract from bg(x)

we get 

        $\begin{array}{rcl}ab{x}^2+2acx+ad-ab{x}^2-2b^{2}x-bc& =& 0\\ 2x\left(ac-b^2\right)& =& bc-ad\\ x& =& \frac{bc-ad}{2\left(ac-b^2\right)}\end{array}$

Multiply the equation (1) with b and subtract $cg\left(x\right)=0$

Hence, 

       $\begin{array}{rcl}{b}^2{x}^2+2bcx+bd-ac{x}^2-2bcx-c^2& =& 0\\ x^2\left(b^2-ac\right)& =& c^2-bd\\ x^2& =& \frac{c^2-bd}{b^2-ac}\end{array}$

From the above unknowns 

$\begin{array}{rcl}{\left[\frac{bc-ad}{2\left(ac-b^2\right)}\right]}^2& =& \frac{c^2-bd}{b^2-ac}\\ {\left(bc-ad\right)}^2& =& 4\left(c^2-bd\right)\left(b^2-ac\right)\end{array}$

Therefore, the required condition is $\begin{array}{rcl} & & \boxed{{\left(bc-ad\right)}^2=4\left(bd-c^2\right)\left(ac-b^2\right)}\end{array}$