 Let $z_1$ and $z_2$ be the zeros of the polynomial $f(x) = x^2 + 6x + 11$. Compute $(1 + z^2_1z_2)(1 + z_1z_2^2)$.
1. Given the polynomial \( f(x) = x^2 + 6x + 11 \), we know that the roots \( z_1 \) and \( z_2 \) satisfy Vieta's formulas:
   \[
   z_1 + z_2 = -\text{coefficient of } x = -6
   \]
   \[
   z_1 z_2 = \text{constant term} = 11
   \]

2. We need to compute \( (1 + z_1^2 z_2)(1 + z_1 z_2^2) \). First, let's expand this expression:
   \[
   (1 + z_1^2 z_2)(1 + z_1 z_2^2) = 1 + z_1^2 z_2 + z_1 z_2^2 + z_1^3 z_2^3
   \]

3. Substitute \( z_1 + z_2 = -6 \) and \( z_1 z_2 = 11 \) into the expanded expression:
   \[
   1 + z_1^2 z_2 + z_1 z_2^2 + z_1^3 z_2^3
   \]

4. Calculate each term separately:
   \[
   z_1^2 z_2 = z_1 (z_1 z_2) = z_1 \cdot 11 = 11z_1
   \]
   \[
   z_1 z_2^2 = z_2 (z_1 z_2) = z_2 \cdot 11 = 11z_2
   \]
   \[
   z_1^3 z_2^3 = (z_1 z_2)^3 = 11^3 = 1331
   \]

5. Substitute these values back into the expression:
   \[
   1 + 11z_1 + 11z_2 + 1331
   \]

6. Use \( z_1 + z_2 = -6 \) to simplify:
   \[
   1 + 11(z_1 + z_2) + 1331 = 1 + 11(-6) + 1331
   \]
   \[
   1 - 66 + 1331 = 1266
   \]

The final answer is \(\boxed{1266}\)