$\Delta=\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|$

Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3},$ we have:

$\Delta=\left|\begin{array}{ccc}5 x+4 & 5 x+4 & 5 x+4 \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|$

$=(5 x+4)\left|\begin{array}{ccc}1 & 0 & 0 \\ 2 x & x+4 & 0 \\ 2 x & 0 & x+4\end{array}\right|$

Applying $\mathrm{C}_{2} \rightarrow C_{2}-C_{1}, \mathrm{C}_{3} \rightarrow C_{3}-C_{1},$ we have

$\Delta=(5 x+4)\left|\begin{array}{ccc}1 & 0 & 0 \\ 2 x & -x+4 & 0 \\ 2 x & 0 & -x+4\end{array}\right|$

$=(5 x+4)(4-x)(4-x)\left|\begin{array}{ccc}1 & 0 & 0 \\ 2 x & 1 & 0 \\ 2 x & 0 & 1\end{array}\right|$

Expanding along $C_{3},$ we have:

$\Delta=(5 x+4)(4-x)^{2}\left|\begin{array}{cc}1 & 0 \\ 2 x & 1\end{array}\right|$

$=(5 x+4)(4-x)^{2}$

Hence, the given result is proved.