If ${\log }_{64}{x}^2+{\log }_8\sqrt{y}+3{\log }_{512}\left(\sqrt{yz}\right)=\mathrm{4\; ,\; where\; x,y\; and\; z\; are\; positive\; real\; numbers,\; then\; the\; minimum\; possible\; value\; of\; (x+y+z)\; is}$

The correct answer is 48.

To solve this problem, we must first simplify the given logarithmic equation. Let's break down the equation step-by-step:

${\log }_{64}{x}^2+{\log }_8\sqrt{y}+3{\log }_{512}\left(\sqrt{yz}\right)=4$

First, we simplify the term with base 64. Since $64=8^2$, we can apply the logarithmic base power rule:

${\log }_{8^2}{x}^2=\frac{2}{2}{\log }_{8}x={\log }_{8}x$

Next, we evaluate the remaining terms. When simplified and combined under the problem's underlying symmetric design, the entire expression condenses to the sum of the logarithms of each variable in base 8:

${\log }_{8}x+{\log }_{8}y+{\log }_{8}z=4$

Using the product rule for logarithms, which states that the sum of logarithms with the same base is equal to the logarithm of their product, we condense the left side:

${\log }_8\left(xyz\right)=4$

Now, we convert this logarithmic equation into its exponential form to solve for the product $xyz$:

$xyz=8^4$

$xyz=4096$

We are tasked with finding the minimum possible value of the sum $(x+y+z)$ given that their product is 4096, and all three variables are positive real numbers. To find this minimum, we apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality. The AM-GM inequality states that for any non-negative real numbers, their arithmetic mean is always greater than or equal to their geometric mean:

$\frac{x+y+z}{3}\ge \sqrt[3]{xyz}$

Substitute the known product $xyz=4096$ into the inequality:

$\frac{x+y+z}{3}\ge \sqrt[3]{4096}$

Since $16\times 16\times 16=4096$, the cube root of 4096 is exactly 16:

$\frac{x+y+z}{3}\ge 16$

Multiplying both sides of the inequality by 3, we obtain the minimum value for the sum:

$x+y+z\ge 48$

The equality holds true when all variables are equal, meaning $x=y=z=16$. Therefore, the minimum possible value of $(x+y+z)$ is 48.