$f\left(x\right)$$=\left|\begin{array}{ccc}x+\lambda & x& x\\ x& x+\lambda & x\\ x& x& x+\lambda \end{array}\right|$then $\mathrm{f}\left(3\mathrm{x}\right)-\mathrm{f}\left(\mathrm{x}\right)=$

 $$\begin{gathered} Given f\left(x\right)=\left|\begin{array}{ccc}x+\lambda & x& x\\ x& x+\lambda & x\\ x& x& x+\lambda \end{array}\right| \\ Apply \left.\begin{array}{l}{\mathrm{R}}_1\to {\mathrm{R}}_1-{\mathrm{R}}_2\\ {\mathrm{R}}_2\to {\mathrm{R}}_2-{\mathrm{R}}_3\end{array}\right\} \\ =\left|\begin{array}{ccc}\lambda & -\lambda & 0\\ 0& \lambda & -\lambda \\ x& x& x+\lambda \end{array}\right| \\ =\lambda (\lambda x+{\lambda }^2+\lambda x)+\lambda (0+\lambda x)+0 \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=3{\mathrm{\lambda }}^2\mathrm{x}+{\mathrm{\lambda }}^3 \end{gathered}$$
$\therefore \mathrm{f}\left(3\mathrm{x}\right)-\mathrm{f}\left(\mathrm{x}\right)=6{\mathrm{\lambda }}^2\mathrm{x}$