Suppose that $ \left(u_n\right)$ is a sequence of real numbers satisfying $ u_{n \plus{} 2} \equal{} 2u_{n \plus{} 1} \plus{} u_{n}$, and that $ u_3 \equal{} 9$ and $ u_6 \equal{} 128$. What is $ u_5$?

$ \textbf{(A)}\ 40 \qquad \textbf{(B)}\ 53 \qquad \textbf{(C)}\ 68 \qquad \textbf{(D)}\ 88 \qquad \textbf{(E)}\ 104$
1. Given the recurrence relation \( u_{n+2} = 2u_{n+1} + u_n \), we need to find \( u_5 \) given \( u_3 = 9 \) and \( u_6 = 128 \).

2. We start by expressing \( u_6 \) in terms of \( u_4 \) and \( u_5 \):
   \[
   u_6 = 2u_5 + u_4
   \]
   Given \( u_6 = 128 \), we have:
   \[
   128 = 2u_5 + u_4 \quad \text{(1)}
   \]

3. Next, we express \( u_5 \) in terms of \( u_3 \) and \( u_4 \):
   \[
   u_5 = 2u_4 + u_3
   \]
   Given \( u_3 = 9 \), we have:
   \[
   u_5 = 2u_4 + 9 \quad \text{(2)}
   \]

4. Substitute equation (2) into equation (1):
   \[
   128 = 2(2u_4 + 9) + u_4
   \]
   Simplify the equation:
   \[
   128 = 4u_4 + 18 + u_4
   \]
   Combine like terms:
   \[
   128 = 5u_4 + 18
   \]
   Solve for \( u_4 \):
   \[
   128 - 18 = 5u_4
   \]
   \[
   110 = 5u_4
   \]
   \[
   u_4 = \frac{110}{5} = 22
   \]

5. Substitute \( u_4 = 22 \) back into equation (2) to find \( u_5 \):
   \[
   u_5 = 2(22) + 9
   \]
   \[
   u_5 = 44 + 9
   \]
   \[
   u_5 = 53
   \]

The final answer is \( \boxed{53} \)