Put $x = 0 \Rightarrow \left| {\begin{array}{*{20}{c}}
{ - 4}&0&0\\
0&{ - 4}&0\\
0&0&{ - 4}
\end{array}} \right| = {A^3} \Rightarrow {A^3} = {\left( { - 4} \right)^3}$

$ \Rightarrow A =  - 4$

$ \Rightarrow \left| {\begin{array}{*{20}{c}}
{x - 4}&{2x}&{2x}\\
{2x}&{x - 4}&{2x}\\
{2x}&{2x}&{x - 4}
\end{array}} \right| = \left( {Bx - 4} \right){\left( {x - 4} \right)^2}$

Now take $x$ common from both the sides

$\therefore \left| {\begin{array}{*{20}{c}}
{1 - \frac{4}{x}}&{2x}&{2x}\\
{2x}&{1 - \frac{4}{x}}&{1 - \frac{4}{x}}\\
{2x}&{2x}&{2x}
\end{array}} \right| = \left( {B - \frac{4}{x}} \right){\left( {1 + \frac{4}{x}} \right)^2}$