First, let's start by substituting $$y = (8)^x$$ in the given equation. By substituting, the equation $$8^{2x} - 16 \cdot 8^x + 48 = 0$$ will be transformed into

$$ y^2 - 16y + 48 = 0 $$

Now, we have a quadratic equation in $$y$$. To find the roots of this quadratic equation, we can use the quadratic formula:

$$ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

In this case, $$a = 1$$, $$b = -16$$, and $$c = 48$$. Substituting these values into the formula, we get:

$$ y = \frac{16 \pm \sqrt{256 - 192}}{2} $$

$$ y = \frac{16 \pm \sqrt{64}}{2} $$

$$ y = \frac{16 \pm 8}{2} $$

Solving for the two possible values of $$y$$, we have:

$$ y = \frac{16 + 8}{2} = 12 $$

$$ y = \frac{16 - 8}{2} = 4 $$

Now, recall that we substituted $$y = (8)^x$$. So, we need to solve for $$x$$ when $$y = 12$$ and $$y = 4$$:

$$ 8^x = 12 $$

$$ x = \log_8(12) $$

$$ 8^x = 4 $$

$$ x = \log_8(4) $$

Therefore, the solutions for $$x$$ are $$\log_8(12)$$ and $$\log_8(4)$$. The sum of these solutions is:

$$ \log_8(12) + \log_8(4) $$

Using the logarithmic property that $$\log_b(m) + \log_b(n) = \log_b(m \cdot n)$$, we get:

$$ \log_8(12) + \log_8(4) = \log_8(12 \cdot 4) $$

$$ = \log_8(48) $$

Now, we note that:

$$ 48 = 8 \cdot 6 $$

Thus,

$$ \log_8(48) = \log_8(8 \cdot 6) = \log_8(8) + \log_8(6) = 1 + \log_8(6) $$

Therefore, the sum of all the solutions of the equation is:

Option A $$1 + \log_8(6)$$.