If  $27a+9b+3c+d=0$ , then the equation  $4a{x}^3+3b{x}^2+2cx+d=0$  has atleast one real root lying between  $\alpha$  and  $\beta$ then  $\alpha +\beta =$_______

$$\begin{gathered} Let f\left(x\right)=\frac{4a{x}^4}{4}+\frac{3b{x}^3}{3}+\frac{2c{x}^2}{2}+dx \\ \Rightarrow f\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx \end{gathered}$$

 $$\begin{gathered} f\left(3\right)=81a+27b+9c+3d \\ =3\left(27a+9b+3c+d\right) \\ \Rightarrow f\left(0\right)=0=f\left(3\right) \\ \mathrm{According} \mathrm{to} \mathrm{Rolles} \mathrm{theorem} \\ \Rightarrow \exists c\in \left(0,3\right)∍f^{\text{'}}\left(c\right)=0 \\ \therefore \alpha =0,\beta =3 \end{gathered}$$