Solve the equation $y^3 = x^3 + 8x^2 - 6x +8$, for positive integers $x$ and $y$.
1. We start with the given equation:
   \[
   y^3 = x^3 + 8x^2 - 6x + 8
   \]
   We need to find positive integer solutions for \(x\) and \(y\).

2. To solve this, we use the technique of bounding. We first consider the expression \((x+2)^3\):
   \[
   (x+2)^3 = x^3 + 6x^2 + 12x + 8
   \]
   We compare this with the given expression \(x^3 + 8x^2 - 6x + 8\).

3. For \(x > 9\), we analyze the inequality:
   \[
   x^3 + 6x^2 + 12x + 8 < x^3 + 8x^2 - 6x + 8
   \]
   Simplifying, we get:
   \[
   6x^2 + 12x < 8x^2 - 6x
   \]
   \[
   0 < 2x^2 - 18x
   \]
   \[
   0 < 2x(x - 9)
   \]
   Since \(x > 9\), the inequality holds true.

4. Next, we consider the expression \((x+3)^3\):
   \[
   (x+3)^3 = x^3 + 9x^2 + 27x + 27
   \]
   We compare this with the given expression \(x^3 + 8x^2 - 6x + 8\).

5. For \(x > 9\), we analyze the inequality:
   \[
   x^3 + 8x^2 - 6x + 8 < x^3 + 9x^2 + 27x + 27
   \]
   Simplifying, we get:
   \[
   8x^2 - 6x + 8 < 9x^2 + 27x + 27
   \]
   \[
   0 < x^2 + 33x + 19
   \]
   This inequality is always true for positive \(x\).

6. Therefore, for \(x > 9\), the expression \(x^3 + 8x^2 - 6x + 8\) lies between \((x+2)^3\) and \((x+3)^3\), which means there are no integer solutions for \(y\) in this range. Hence, our assumption that \(x > 9\) is false.

7. We now check for values of \(x \leq 9\):
   - For \(x = 1\):
     \[
     y^3 = 1^3 + 8(1)^2 - 6(1) + 8 = 1 + 8 - 6 + 8 = 11 \quad (\text{not a perfect cube})
     \]
   - For \(x = 2\):
     \[
     y^3 = 2^3 + 8(2)^2 - 6(2) + 8 = 8 + 32 - 12 + 8 = 36 \quad (\text{not a perfect cube})
     \]
   - For \(x = 3\):
     \[
     y^3 = 3^3 + 8(3)^2 - 6(3) + 8 = 27 + 72 - 18 + 8 = 89 \quad (\text{not a perfect cube})
     \]
   - For \(x = 4\):
     \[
     y^3 = 4^3 + 8(4)^2 - 6(4) + 8 = 64 + 128 - 24 + 8 = 176 \quad (\text{not a perfect cube})
     \]
   - For \(x = 5\):
     \[
     y^3 = 5^3 + 8(5)^2 - 6(5) + 8 = 125 + 200 - 30 + 8 = 303 \quad (\text{not a perfect cube})
     \]
   - For \(x = 6\):
     \[
     y^3 = 6^3 + 8(6)^2 - 6(6) + 8 = 216 + 288 - 36 + 8 = 476 \quad (\text{not a perfect cube})
     \]
   - For \(x = 7\):
     \[
     y^3 = 7^3 + 8(7)^2 - 6(7) + 8 = 343 + 392 - 42 + 8 = 701 \quad (\text{not a perfect cube})
     \]
   - For \(x = 8\):
     \[
     y^3 = 8^3 + 8(8)^2 - 6(8) + 8 = 512 + 512 - 48 + 8 = 984 \quad (\text{not a perfect cube})
     \]
   - For \(x = 9\):
     \[
     y^3 = 9^3 + 8(9)^2 - 6(9) + 8 = 729 + 648 - 54 + 8 = 1331 = 11^3 \quad (\text{perfect cube})
     \]
     Thus, \(y = 11\).

8. Therefore, the only solution for positive integers \(x\) and \(y\) is:
   \[
   (x, y) = (9, 11)
   \]

The final answer is \(\boxed{(x, y) = (9, 11)}\)