Find all triples $(a,b,c)$ of positive real numbers satisfying the system of equations
\[ a\sqrt{b}-c \&= a,\qquad b\sqrt{c}-a \&= b,\qquad c\sqrt{a}-b \&= c.  \]
We are given the system of equations:
\[
\begin{cases}
a\sqrt{b} - c = a \\
b\sqrt{c} - a = b \\
c\sqrt{a} - b = c
\end{cases}
\]
and we need to find all triples \((a, b, c)\) of positive real numbers that satisfy these equations.

1. **Rewriting the equations:**
   \[
   a\sqrt{b} - c = a \implies a\sqrt{b} = a + c \implies \sqrt{b} = 1 + \frac{c}{a}
   \]
   \[
   b\sqrt{c} - a = b \implies b\sqrt{c} = a + b \implies \sqrt{c} = 1 + \frac{a}{b}
   \]
   \[
   c\sqrt{a} - b = c \implies c\sqrt{a} = b + c \implies \sqrt{a} = 1 + \frac{b}{c}
   \]

2. **Assuming \(a = b = c\):**
   Let's assume \(a = b = c\). Then the equations become:
   \[
   a\sqrt{a} - a = a \implies a\sqrt{a} = 2a \implies \sqrt{a} = 2 \implies a = 4
   \]
   \[
   b\sqrt{b} - b = b \implies b\sqrt{b} = 2b \implies \sqrt{b} = 2 \implies b = 4
   \]
   \[
   c\sqrt{c} - c = c \implies c\sqrt{c} = 2c \implies \sqrt{c} = 2 \implies c = 4
   \]
   Therefore, \((a, b, c) = (4, 4, 4)\) is a solution.

3. **Checking if there are other solutions:**
   Suppose \(a, b, c \neq 4\). By the pigeonhole principle, two of the numbers must either both be greater than or both be less than 4. Let's analyze these cases:

   - **Case 1: \(a, b > 4\):**
     \[
     a\sqrt{b} - c = a \implies a\sqrt{b} = a + c \implies c = a(\sqrt{b} - 1)
     \]
     Since \(a > 4\) and \(\sqrt{b} > 2\), we have \(c > a\). Similarly, we can show \(b > c\) and \(a > b\), leading to a contradiction.

   - **Case 2: \(a, b < 4\):**
     \[
     a\sqrt{b} - c = a \implies a\sqrt{b} = a + c \implies c = a(\sqrt{b} - 1)
     \]
     Since \(a < 4\) and \(\sqrt{b} < 2\), we have \(c < a\). Similarly, we can show \(b < c\) and \(a < b\), leading to a contradiction.

Therefore, the only possible solution is \((a, b, c) = (4, 4, 4)\).

The final answer is \( \boxed{ (4, 4, 4) } \).