If $ x \equal{} \sqrt [3]{11 \plus{} \sqrt {337}} \plus{} \sqrt [3]{11 \minus{} \sqrt {337}}$, then $ x^3 \plus{} 18x$ = ?

$\textbf{(A)}\ 24  \qquad\textbf{(B)}\ 22  \qquad\textbf{(C)}\ 20  \qquad\textbf{(D)}\ 11  \qquad\textbf{(E)}\ 10$
1. Let \( a = \sqrt[3]{11 + \sqrt{337}} \) and \( b = \sqrt[3]{11 - \sqrt{337}} \). We are given that \( x = a + b \).

2. We need to find \( x^3 + 18x \). To do this, we start by considering the expression for \( x^3 \):
   \[
   x^3 = (a + b)^3
   \]
   Expanding the right-hand side using the binomial theorem, we get:
   \[
   x^3 = a^3 + b^3 + 3ab(a + b)
   \]

3. We know that:
   \[
   a^3 = 11 + \sqrt{337} \quad \text{and} \quad b^3 = 11 - \sqrt{337}
   \]
   Adding these two equations, we get:
   \[
   a^3 + b^3 = (11 + \sqrt{337}) + (11 - \sqrt{337}) = 22
   \]

4. Next, we need to find \( ab \). Since \( a \) and \( b \) are cube roots, we can multiply them:
   \[
   ab = \sqrt[3]{(11 + \sqrt{337})(11 - \sqrt{337})}
   \]
   Using the difference of squares, we get:
   \[
   (11 + \sqrt{337})(11 - \sqrt{337}) = 11^2 - (\sqrt{337})^2 = 121 - 337 = -216
   \]
   Therefore:
   \[
   ab = \sqrt[3]{-216} = -6
   \]

5. Substituting \( a^3 + b^3 \) and \( ab \) back into the expression for \( x^3 \), we get:
   \[
   x^3 = 22 + 3(-6)(a + b)
   \]
   Since \( x = a + b \), we can substitute \( x \) for \( a + b \):
   \[
   x^3 = 22 + 3(-6)x = 22 - 18x
   \]

6. Finally, we need to find \( x^3 + 18x \):
   \[
   x^3 + 18x = (22 - 18x) + 18x = 22
   \]

The final answer is \(\boxed{22}\).