Find every real values that $a$ can assume such that
$$\begin{cases}
x^3 + y^2 + z^2 = a\\
x^2 + y^3 + z^2 = a\\
x^2 + y^2 + z^3 = a
\end{cases}$$
has a solution with $x, y, z$ distinct real numbers.
To find the real values that \( a \) can assume such that the system of equations
\[
\begin{cases}
x^3 + y^2 + z^2 = a \\
x^2 + y^3 + z^2 = a \\
x^2 + y^2 + z^3 = a
\end{cases}
\]
has a solution with \( x, y, z \) being distinct real numbers, we proceed as follows:

1. **Compare the equations:**
   By comparing the given equations, we can write:
   \[
   x^3 + y^2 + z^2 = x^2 + y^3 + z^2 = x^2 + y^2 + z^3 = a
   \]
   This implies:
   \[
   x^3 - x^2 = y^3 - y^2 = z^3 - z^2 = K
   \]
   for some constant \( K \).

2. **Define the polynomial:**
   Let \( P(t) = t^3 - t^2 - K \). Clearly, \( x, y, z \) are the roots of this polynomial. 

3. **Use Vieta's formulas:**
   By Vieta's formulas for the roots of the polynomial \( P(t) = t^3 - t^2 - K \), we have:
   \[
   x + y + z = 1
   \]
   \[
   xy + yz + zx = 0
   \]
   \[
   xyz = -K
   \]

4. **Calculate \( x^2 + y^2 + z^2 \):**
   Using the identity \( (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) \), we get:
   \[
   1^2 = x^2 + y^2 + z^2 + 2 \cdot 0 \implies x^2 + y^2 + z^2 = 1
   \]

5. **Express \( x^3 + y^3 + z^3 \):**
   Since \( x^3 = x^2 + K \), \( y^3 = y^2 + K \), and \( z^3 = z^2 + K \), we have:
   \[
   x^3 + y^3 + z^3 = (x^2 + y^2 + z^2) + 3K = 1 + 3K
   \]

6. **Relate to \( a \):**
   From the given system, we know:
   \[
   3a = (x^3 + y^3 + z^3) + 2(x^2 + y^2 + z^2)
   \]
   Substituting the values, we get:
   \[
   3a = (1 + 3K) + 2 \cdot 1 = 1 + 3K + 2 = 3 + 3K
   \]
   \[
   3a = 3 + 3K \implies a = 1 + K
   \]

7. **Determine the range of \( K \):**
   The polynomial \( P(t) = t^3 - t^2 - K \) has three distinct real roots if and only if its discriminant is positive. The discriminant of \( t^3 - t^2 - K \) is positive if \( K \in \left(-\frac{4}{27}, 0\right) \).

8. **Find the range of \( a \):**
   Since \( a = 1 + K \) and \( K \in \left(-\frac{4}{27}, 0\right) \), we have:
   \[
   a \in \left(1 - \frac{4}{27}, 1\right) = \left(\frac{23}{27}, 1\right)
   \]

Thus, the values that \( a \) can assume are in the interval \(\left(\frac{23}{27}, 1\right)\).

The final answer is \(\boxed{\left(\frac{23}{27}, 1\right)}\).