To find the minimum value of the expression
$$A = \frac{2-a^3}{a} + \frac{2-b^3}{b} + \frac{2-c^3}{c},$$
given that $a, b, c$ are positive real numbers and $a + b + c = 3$, we proceed as follows:

First, we rewrite the expression:
$$A = \frac{2}{a} - a^2 + \frac{2}{b} - b^2 + \frac{2}{c} - c^2.$$

Consider the function of a single variable:
$$f(x) = \frac{2}{x} - x^2.$$
Our goal is to minimize the sum $f(a) + f(b) + f(c)$.

We calculate the derivative of $f(x)$:
$$f'(x) = -\frac{2}{x^2} - 2x.$$
Setting $f'(x) = 0$ to find critical points:
$$-\frac{2}{x^2} - 2x = 0 \quad \Rightarrow \quad \frac{2}{x^2} = -2x \quad \Rightarrow \quad x^3 = 1.$$
Thus, $x = 1$ is a critical point where $f(x)$ is extremized.

Checking the nature of this critical point, we compute the second derivative:
$$f''(x) = \frac{4}{x^3} - 2.$$
Evaluating $f''(1)$:
$$f''(1) = 4 - 2 = 2 > 0.$$
This indicates that $x = 1$ is a point of local minimum for $f(x)$.

Considering the condition $a + b + c = 3$, a natural symmetry suggests $a = b = c = 1$. Substituting these values back into the expression for $A$:
$$A = \left( \frac{2}{1} - 1^2 \right) + \left( \frac{2}{1} - 1^2 \right) + \left( \frac{2}{1} - 1^2 \right) = (2 - 1) + (2 - 1) + (2 - 1) = 3.$$

Therefore, the minimum value of $A$ is:
$$\boxed{3}.$$
This minimum is achieved when $a = b = c = 1$.