Find all positive integer bases $b \ge 9$ so that the number 
\[ \frac{{\overbrace{11 \cdots 1}^{n-1 \ 1's}0\overbrace{77 \cdots 7}^{n-1\ 7's}8\overbrace{11 \cdots 1}^{n \ 1's}}_b}{3} \]
is a perfect cube in base 10 for all sufficiently large positive integers $n$.

[i]Proposed by Yang Liu[/i]
1. **Express the number in base \( b \)**:
   The given number in base \( b \) can be written as:
   \[
   \frac{\overbrace{11 \cdots 1}^{n-1 \text{ 1's}}0\overbrace{77 \cdots 7}^{n-1 \text{ 7's}}8\overbrace{11 \cdots 1}^{n \text{ 1's}}_b}{3}
   \]
   This number can be expanded as:
   \[
   \frac{b^{3n} - b^{2n+1} + 7 \cdot b^{2n} + b^{n+1} - 7 \cdot b^n - 1}{3(b-1)}
   \]

2. **Simplify the expression**:
   For large \( n \), the dominant term in the numerator is \( b^{3n} \). Therefore, the expression can be approximated by:
   \[
   \frac{b^{3n} - b^{2n+1} + 7 \cdot b^{2n} + b^{n+1} - 7 \cdot b^n - 1}{3(b-1)}
   \]

3. **Approximate the cube root**:
   The cube root of the expression for large \( n \) is approximately:
   \[
   \frac{b^n - \frac{b-7}{3}}{(3(b-1))^{\frac{1}{3}}}
   \]
   This approximation suggests that \( 3(b-1) \) must be a perfect cube.

4. **Determine the form of \( b \)**:
   Since \( 3(b-1) \) must be a perfect cube, let:
   \[
   3(b-1) = k^3
   \]
   for some integer \( k \). Solving for \( b \), we get:
   \[
   b = \frac{k^3}{3} + 1
   \]

5. **Check for integer solutions**:
   For \( b \) to be an integer, \( k^3 \) must be divisible by 3. Let \( k = 3m \) for some integer \( m \). Then:
   \[
   b = \frac{(3m)^3}{3} + 1 = 9m^3 + 1
   \]

6. **Verify the solution**:
   Plugging \( b = 9m^3 + 1 \) back into the original expression, we need to check if the number is a perfect cube for all sufficiently large \( n \). The simplest case is when \( m = 1 \), giving \( b = 10 \).

   For \( b = 10 \):
   \[
   \frac{10^{3n} - 10^{2n+1} + 7 \cdot 10^{2n} + 10^{n+1} - 7 \cdot 10^n - 1}{3 \cdot 9}
   \]
   Simplifies to a perfect cube for large \( n \).

   Therefore, \( b = 10 \) is a valid solution.

The final answer is \( \boxed{ b = 10 } \)