If the expression ${\mathrm{x}}^3+3{\mathrm{x}}^2-9\mathrm{x}+\mathrm{\lambda }$ is of the form $(\mathrm{x}-\mathrm{\alpha }{)}^2(\mathrm{x}-\mathrm{\beta })$, then the values of $\mathrm{\lambda }$ are

$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3+3{\mathrm{x}}^2-9\mathrm{x}+\mathrm{\lambda }$

By observation of $(\mathrm{x}-\mathrm{\alpha }{)}^2(\mathrm{x}-\mathrm{\beta })$ is $\alpha$ is a double root so, $\alpha$ is also root of $f^{\text{'}}\left(x\right)=0$ then

${\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=3x^2+6x-9$

$$\begin{gathered} 3x^2+6x-9=0 \\ {x}^2+2x-3=0 \\ \left(x-1\right)\left(x+3\right)=0 \\ x=1,-3 \end{gathered}$$

$If x=1 then$

$\begin{array}{r}\Rightarrow 1+3-9+c=0\\ \Rightarrow \lambda -5=0\\ \Rightarrow \lambda =5\\ \text{ If x}=3\text{ then, }\\ \Rightarrow -27+27+27+\lambda =0\\ \Rightarrow \lambda =-27\end{array}$