Prove the following identities

LHS :

$\left|\begin{array}{ccc}x+\lambda & 2 x & 2 x \\ 2 x & x+\lambda & 2 x \\ 2 x & 2 x & x+\lambda\end{array}\right|$

$=\left|\begin{array}{ccc}x+\lambda & 2 x & 2 x \\ 2 x-x-\lambda & x+\lambda-2 x & 0 \\ 2 x-x-\lambda & 0 & x+\lambda-2 x\end{array}\right| \quad$ [Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$ ]

$=\left|\begin{array}{ccc}x+\lambda & 2 x & 2 x \\ -(\lambda-x) & \lambda-x & 0 \\ -(\lambda-x) & 0 & \lambda-x\end{array}\right|$

$=(\lambda-x)^{2}\left|\begin{array}{ccc}x+\lambda & 2 x & 2 x \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right|$              $\left[\right.$ Taking $(\lambda-x)$ common from $R_{2}$ and $(\lambda-x)$ common from $\left.R_{3}\right]$

$=(\lambda-x)^{2}[-1(-2 x)+1(x+\lambda+2 x)] \quad$ [Expanding along last row]

$=(\lambda-x)^{2}(\lambda+5 x)$

$=$ RHS

$\therefore\left|\begin{array}{ccc}x+\lambda & 2 x & 2 x \\ 2 x & x+\lambda & 2 x \\ 2 x & 2 x & x+\lambda\end{array}\right|=(\lambda-x)^{2}(\lambda+5 x)$