Let $a, b, c$ be the roots of the equation $x^3-9x^2+11x-1 = 0$, and define $s =\sqrt{a}+\sqrt{b}+\sqrt{c}$. 
Compute $s^4 -18s^2 - 8s$ .
1. Given the polynomial equation \(x^3 - 9x^2 + 11x - 1 = 0\), we know from Vieta's formulas that the roots \(a, b, c\) satisfy:
   \[
   a + b + c = 9, \quad ab + bc + ca = 11, \quad abc = 1.
   \]

2. Define \(s = \sqrt{a} + \sqrt{b} + \sqrt{c}\). We start by squaring \(s\):
   \[
   s^2 = (\sqrt{a} + \sqrt{b} + \sqrt{c})^2 = a + b + c + 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}).
   \]
   Using \(a + b + c = 9\), we get:
   \[
   s^2 = 9 + 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}).
   \]

3. Isolate the term involving the square roots:
   \[
   s^2 - 9 = 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}).
   \]

4. Square both sides to eliminate the square roots:
   \[
   (s^2 - 9)^2 = \left(2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca})\right)^2.
   \]
   Simplifying the right-hand side:
   \[
   (s^2 - 9)^2 = 4(\sqrt{ab} + \sqrt{bc} + \sqrt{ca})^2.
   \]

5. Expand the right-hand side:
   \[
   (\sqrt{ab} + \sqrt{bc} + \sqrt{ca})^2 = ab + bc + ca + 2(\sqrt{a^2bc} + \sqrt{ab^2c} + \sqrt{abc^2}).
   \]
   Using \(ab + bc + ca = 11\) and \(abc = 1\), we get:
   \[
   (\sqrt{ab} + \sqrt{bc} + \sqrt{ca})^2 = 11 + 2(\sqrt{a^2bc} + \sqrt{ab^2c} + \sqrt{abc^2}) = 11 + 2(\sqrt{a} + \sqrt{b} + \sqrt{c}).
   \]

6. Substitute back into the equation:
   \[
   (s^2 - 9)^2 = 4(11 + 2s).
   \]
   Simplify:
   \[
   s^4 - 18s^2 + 81 = 44 + 8s.
   \]

7. Rearrange to find the desired expression:
   \[
   s^4 - 18s^2 - 8s = 44 - 81 = -37.
   \]

The final answer is \(\boxed{-37}\).