Just think different 3

We note that for $x \le 0$ , $\log_2 x^3$ and $\log_x 4$ in the LHS of the equation are undefined. For $0 < x < 2$ , the RHS $<0$ but the LHS $> 0$ , therefore, there is no real root in $0 < x < 2$ . For $x=0$ , the LHS $= 0^{-3}$ which is undefined. Therefore, the real roots are in $x > 2$ . Then we have:

$\begin{aligned} \color{#3D99F6}{|x-2|}^{\log_2 x^3-3 \log_x 4} & = (x-2)^3 \quad \quad \small \color{#3D99F6}{\text{For } x > 2, \space |x-2| = x - 2} \\ \color{#3D99F6}{(x-2)}^{\log_2 x^3-3 \log_x 4} & = (x-2)^3 \quad \Rightarrow \begin{cases} x - 2 = 1 \quad \Rightarrow x = 3 & ...(1) \\ \log_2 x^3-3 \log_x 4 = 3 & ...(2) \end{cases} \end{aligned}$

$\begin{aligned} (2): \quad \log_2 x^3-3 \log_x 4 & = 3 \\ 3 \log_2 x - 3 \frac{\log_2 2^2}{\log_2 x} & = 3 \\ \log_2 x - \frac{2}{\log_2 x} & = 1 \\ \log_2^2 x - \log_2 x - 2 & = 0 \\ (\log_2 x -2)(\log_2 x +1) & = 0 \\ \Rightarrow \log_2 x & = 2 \quad \quad \small \color{#3D99F6}{\log_2 x > 0} \\ \Rightarrow x & = 4 \end{aligned}$

Therefore, the sum of roots $= 3+4 = \boxed{7}$