If $$\log_{64}{x^{2}+\log_{8}{\sqrt{y}+3\log_{512}{(\sqrt{y}z)}}}=4$$, where x,y and z are positive real numbers, then the minimum possible value of (x+y+z) is

$$64 = 8^2  \text{and}  512 = 8^3$$

$$\log_{64}{x^{2}+\log_{8}{\sqrt{y}+3\log_{512}{(\sqrt{y}z)}}}=4$$,

Property of log: $$\log_{b^m}\ a^{n\ }=\frac{n}{m}\ \log_ba$$

$$\log_{8^2}{x^{2}+\log_{8}{\sqrt{y}+3\log_{8^3}{(\sqrt{y}z)}}}=4$$

Using the above-mentioned property, the expression becomes $$\log_{8}{x}+\log_{8}{\sqrt{y}+\log_{8}{(\sqrt{y}z)}}=4$$

$$\log_8x\sqrt{y}\cdot(\sqrt{y}z)=4$$

$$\log_8xyz=4$$

$$xyz =8^4=2^{12}$$

Using AM-GM inequality

$$\frac{\left(x+y+z\right)}{3}\ge\sqrt[\ 3]{xyz}$$

$$\frac{\left(x+y+z\right)}{3}\ge2^4$$

$$\left(x+y+z\right)\ge48$$