Let $a_1, a_2, \ldots$ and $b_1, b_2, \ldots$ be sequences such that $a_ib_i - a_i - b_i = 0$ and $a_{i+1} = \frac{2-a_ib_i}{1-b_i}$ for all $i \ge 1$.  If $a_1 = 1 + \frac{1}{\sqrt[4]{2}}$, then what is $b_{6}$?

[i]Proposed by Andrew Wu[/i]
1. Given the sequences \(a_1, a_2, \ldots\) and \(b_1, b_2, \ldots\) such that \(a_i b_i - a_i - b_i = 0\) and \(a_{i+1} = \frac{2 - a_i b_i}{1 - b_i}\) for all \(i \ge 1\), we start by rearranging the first condition:
   \[
   a_i b_i - a_i - b_i = 0 \implies a_i b_i = a_i + b_i
   \]
   This can be rewritten as:
   \[
   a_i b_i = a_i + b_i \implies (1 - a_i)(1 - b_i) = 1
   \]

2. Using the second condition \(a_{i+1} = \frac{2 - a_i b_i}{1 - b_i}\), we substitute \(a_i b_i = a_i + b_i\):
   \[
   a_{i+1} = \frac{2 - (a_i + b_i)}{1 - b_i}
   \]
   Simplifying further:
   \[
   a_{i+1} = \frac{2 - a_i - b_i}{1 - b_i}
   \]

3. From the rearranged first condition \((1 - a_i)(1 - b_i) = 1\), we get:
   \[
   1 - b_i = \frac{1}{1 - a_i}
   \]
   Substituting this into the expression for \(a_{i+1}\):
   \[
   a_{i+1} = \frac{2 - a_i - b_i}{\frac{1}{1 - a_i}} = (2 - a_i - b_i)(1 - a_i)
   \]

4. Since \(a_i b_i = a_i + b_i\), we can express \(b_i\) in terms of \(a_i\):
   \[
   b_i = \frac{a_i}{a_i - 1}
   \]

5. Now, we need to find the sequence \(a_i\). Given \(a_1 = 1 + \frac{1}{\sqrt[4]{2}}\), we use the recurrence relation:
   \[
   a_{i+1} = 1 + (1 - a_i)^2
   \]

6. Calculate the first few terms of the sequence \(a_i\):
   \[
   a_1 = 1 + \frac{1}{\sqrt[4]{2}}
   \]
   \[
   a_2 = 1 + \left(1 - \left(1 + \frac{1}{\sqrt[4]{2}}\right)\right)^2 = 1 + \left(-\frac{1}{\sqrt[4]{2}}\right)^2 = 1 + \frac{1}{\sqrt{2}}
   \]
   \[
   a_3 = 1 + \left(1 - \left(1 + \frac{1}{\sqrt{2}}\right)\right)^2 = 1 + \left(-\frac{1}{\sqrt{2}}\right)^2 = 1 + \frac{1}{2}
   \]
   \[
   a_4 = 1 + \left(1 - \left(1 + \frac{1}{2}\right)\right)^2 = 1 + \left(-\frac{1}{2}\right)^2 = 1 + \frac{1}{4}
   \]
   \[
   a_5 = 1 + \left(1 - \left(1 + \frac{1}{4}\right)\right)^2 = 1 + \left(-\frac{1}{4}\right)^2 = 1 + \frac{1}{16}
   \]
   \[
   a_6 = 1 + \left(1 - \left(1 + \frac{1}{16}\right)\right)^2 = 1 + \left(-\frac{1}{16}\right)^2 = 1 + \frac{1}{256}
   \]

7. Finally, we find \(b_6\):
   \[
   1 - b_6 = \frac{1}{1 - a_6} = \frac{1}{1 - \left(1 + \frac{1}{256}\right)} = \frac{1}{-\frac{1}{256}} = -256
   \]
   \[
   b_6 = 1 + 256 = 257
   \]

The final answer is \(\boxed{257}\)