Find all triples of positive integers $(a,b,c)$ such that the following equations are both true:
I-  $a^2+b^2=c^2$
II- $a^3+b^3+1=(c-1)^3$
1. **Assume without loss of generality that \(a \leq b\).**

2. **Given the equations:**
   \[
   \text{I. } a^2 + b^2 = c^2
   \]
   \[
   \text{II. } a^3 + b^3 + 1 = (c-1)^3
   \]

3. **We view \(c\) as constant and \(a, b\) as real variables. Under the condition \(a^2 + b^2 = c^2\), it is easy to show that the value of \(a^3 + b^3 + 1\) is an increasing function of \(b\).**

4. **Since \(b\) is an integer and strictly less than \(c - 1\), we have \(b \leq c - 2\). Hence:**
   \[
   (c - 1)^3 = a^3 + b^3 + 1 \leq \sqrt{4c - 4}^3 + (c - 2)^3 + 1
   \]

5. **Simplify the inequality:**
   \[
   (c - 1)^3 \leq \sqrt{4c - 4}^3 + (c - 2)^3 + 1
   \]

6. **To find the upper bound for \(c\), we need to solve:**
   \[
   3(c - 2) \leq 8\sqrt{c - 1}
   \]

7. **Solve the inequality \(3(c - 2) \leq 8\sqrt{c - 1}\):**
   \[
   3c - 6 \leq 8\sqrt{c - 1}
   \]
   \[
   3c - 6 \leq 8\sqrt{c - 1}
   \]
   \[
   \left(\frac{3c - 6}{8}\right)^2 \leq c - 1
   \]
   \[
   \frac{(3c - 6)^2}{64} \leq c - 1
   \]
   \[
   (3c - 6)^2 \leq 64(c - 1)
   \]
   \[
   9c^2 - 36c + 36 \leq 64c - 64
   \]
   \[
   9c^2 - 100c + 100 \leq 0
   \]

8. **Solve the quadratic inequality \(9c^2 - 100c + 100 \leq 0\):**
   \[
   c = \frac{100 \pm \sqrt{10000 - 3600}}{18}
   \]
   \[
   c = \frac{100 \pm \sqrt{6400}}{18}
   \]
   \[
   c = \frac{100 \pm 80}{18}
   \]
   \[
   c = \frac{180}{18} = 10 \quad \text{or} \quad c = \frac{20}{18} \approx 1.11
   \]

   Since \(c\) must be a positive integer, we have \(c \leq 10\).

9. **Perform an exhaustive search for \(c \leq 10\):**
   - For \(c = 10\):
     \[
     a^2 + b^2 = 10^2 = 100
     \]
     Possible pairs \((a, b)\) are \((6, 8)\) and \((8, 6)\) since:
     \[
     6^2 + 8^2 = 36 + 64 = 100
     \]
     Check the second equation:
     \[
     a^3 + b^3 + 1 = (10 - 1)^3 = 9^3 = 729
     \]
     \[
     6^3 + 8^3 + 1 = 216 + 512 + 1 = 729
     \]
     Both pairs \((6, 8, 10)\) and \((8, 6, 10)\) satisfy both equations.

10. **Verify other values of \(c\) (from 1 to 9):**
    - For \(c = 9\):
      \[
      a^2 + b^2 = 81
      \]
      No integer pairs \((a, b)\) satisfy both equations.
    - Similarly, check for \(c = 8, 7, 6, 5, 4, 3, 2, 1\) and find no solutions.

Conclusion:
The only solutions are \((a, b, c) = (6, 8, 10)\) and \((8, 6, 10)\).

The final answer is \( \boxed{ (6, 8, 10) } \) and \((8, 6, 10)\).