Using properties of determinants prove that

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

$=\left|\begin{array}{ccc}5 x+4 & 5 x+4 & 5 x+4 \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|$         [Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$ ]

$=5 x+4\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|$        [Take out $5 x+4$ common from $R_{1}$ ]

$=5 x+4\left|\begin{array}{ccc}1 & 0 & 0 \\ 2 x & 4-x & 0 \\ 2 x & 0 & 4-x\end{array}\right|$       [Applying $C_{2} \rightarrow C_{2}-C_{1}$ and $C_{3} \rightarrow C_{3}-C_{1}$ ]

$=5 x+4(4-x)^{2}$      [Expanding along $R_{1}$ ]

Hence proved.