Show that the number of positive integer solutions to $x_1 + 2^3x_2 + 3^3x_3 + \ldots + 10^3x_{10} = 3025$ (*) equals the number of non-negative integer solutions to the equation $y_1 + 2^3y_2 + 3^3y_3 + \ldots + 10^3y_{10} = 0$. Hence show that (*) has a unique solution in positive integers and find it. 
1. **Understanding the given equation:**
   We are given the equation:
   \[
   x_1 + 2^3x_2 + 3^3x_3 + \ldots + 10^3x_{10} = 3025
   \]
   We need to show that the number of positive integer solutions to this equation equals the number of non-negative integer solutions to the equation:
   \[
   y_1 + 2^3y_2 + 3^3y_3 + \ldots + 10^3y_{10} = 0
   \]

2. **Sum of cubes of the first 10 natural numbers:**
   First, we calculate the sum of the cubes of the first 10 natural numbers:
   \[
   \sum_{i=1}^{10} i^3 = 1^3 + 2^3 + 3^3 + \ldots + 10^3
   \]
   Using the formula for the sum of cubes:
   \[
   \left( \sum_{i=1}^{10} i \right)^2 = \left( \frac{10 \cdot 11}{2} \right)^2 = 55^2 = 3025
   \]
   Therefore,
   \[
   1^3 + 2^3 + 3^3 + \ldots + 10^3 = 3025
   \]

3. **Positive integer solutions to the first equation:**
   Since the sum of the cubes of the first 10 natural numbers is 3025, we can rewrite the given equation as:
   \[
   x_1 \cdot 1^3 + x_2 \cdot 2^3 + x_3 \cdot 3^3 + \ldots + x_{10} \cdot 10^3 = 3025
   \]
   Given that the sum of the cubes is exactly 3025, and each term \(i^3\) is unique and positive, the only way to achieve this sum with positive integer coefficients \(x_i\) is to have each \(x_i = 1\). Therefore, the unique solution in positive integers is:
   \[
   x_1 = x_2 = x_3 = \ldots = x_{10} = 1
   \]

4. **Non-negative integer solutions to the second equation:**
   For the equation:
   \[
   y_1 + 2^3y_2 + 3^3y_3 + \ldots + 10^3y_{10} = 0
   \]
   Since all terms on the left-hand side are non-negative and the right-hand side is zero, the only solution is:
   \[
   y_1 = y_2 = y_3 = \ldots = y_{10} = 0
   \]

5. **Conclusion:**
   Both equations have exactly one solution. The first equation has the unique positive integer solution:
   \[
   x_1 = x_2 = x_3 = \ldots = x_{10} = 1
   \]
   The second equation has the unique non-negative integer solution:
   \[
   y_1 = y_2 = y_3 = \ldots = y_{10} = 0
   \]

The final answer is \( \boxed{ x_1 = x_2 = x_3 = \ldots = x_{10} = 1 } \)