Find all triples $(a,b,c)$ of real numbers all different from zero that satisfies: 

\begin{eqnarray}  a^4+b^2c^2=16a\nonumber \\ b^4+c^2a^2=16b \nonumber\\ c^4+a^2b^2=16c \nonumber \end{eqnarray}

To find all triples \((a, b, c)\) of real numbers, all different from zero, that satisfy the given system of equations:

\[
\begin{cases}
a^4 + b^2c^2 = 16a \\
b^4 + c^2a^2 = 16b \\
c^4 + a^2b^2 = 16c
\end{cases}
\]

we will analyze the problem step by step.

1. **Assume \(a = b = c\):**

   If \(a = b = c\), then substituting into the equations, we get:
   \[
   a^4 + a^4 = 16a \implies 2a^4 = 16a \implies a^4 = 8a \implies a(a^3 - 8) = 0
   \]
   Since \(a \neq 0\), we have:
   \[
   a^3 = 8 \implies a = 2
   \]
   Therefore, \(a = b = c = 2\) is a solution. The triple \((2, 2, 2)\) satisfies all three equations.

2. **Assume \(a > b = c\):**

   Let's consider \(a > b = c\). Substituting \(b = c\) into the equations, we get:
   \[
   a^4 + b^4 = 16a \quad \text{(Equation 1)}
   \]
   \[
   b^4 + a^2b^2 = 16b \quad \text{(Equation 2)}
   \]
   Subtracting Equation 2 from Equation 1:
   \[
   a^4 - a^2b^2 = 16a - 16b \implies a^2(a^2 - b^2) = 16(a - b)
   \]
   Factoring the left-hand side:
   \[
   a^2(a - b)(a + b) = 16(a - b)
   \]
   Since \(a \neq b\), we can divide both sides by \(a - b\):
   \[
   a^2(a + b) = 16
   \]
   Since \(a > b\), we have \(a^2(a + b) > 2b^3\). Then:
   \[
   b < 2
   \]
   Inspecting Equation 2:
   \[
   b^3 + a^2b = 16 > 8 + 2a^2
   \]
   Then:
   \[
   2 > a
   \]
   But:
   \[
   2a^3 > a^2(a + b) = 16 \implies a > 2
   \]
   This is a contradiction. Therefore, there are no solutions in this case.

3. **Assume \(a > b > c\):**

   Let's consider \(a > b > c\). We have the equations:
   \[
   a^4 + b^2c^2 = 16a \quad \text{(Equation X)}
   \]
   \[
   b^4 + c^2a^2 = 16b \quad \text{(Equation Y)}
   \]
   \[
   c^4 + a^2b^2 = 16c \quad \text{(Equation Z)}
   \]
   Subtracting \(2 \times \text{(Equation Y)}\) and \(\text{(Equation Z)}\) from \(\text{(Equation X)}\):
   \[
   a^4 + b^2c^2 - 2(b^4 + c^2a^2) - (c^4 + a^2b^2) = 16a - 32b - 16c
   \]
   Simplifying:
   \[
   a^4 - 2b^4 - c^4 + b^2c^2 - 2c^2a^2 - a^2b^2 = 16a - 32b - 16c
   \]
   This equation is complex and does not simplify easily. However, we can analyze the inequality:
   \[
   a((a + b)^2 + ac) = c((b + c)^2 + ac)
   \]
   Since \(a > c\), we need to prove:
   \[
   (a + b)^2 < (b + c)^2
   \]
   This is incorrect, leading to a contradiction.

Therefore, the only solution is when \(a = b = c = 2\).

The final answer is \( \boxed{ (2, 2, 2) } \).