Is the question complete?

For example, if we plug in a = 1, then we will get (possibly complex) solutions of b.

Computation: 

if a = 1

\( (1-b)(1 - \sqrt{b} ) = 1 \)

Expanding we get \( 1 - \sqrt{b} - b + b \sqrt{b} = 1 \)

This implies \( b \sqrt b = b + \sqrt b \Rightarrow b = \sqrt b + 1 \)

Set \( \sqrt b =x \) and solve the qudratic.

This process will work for any value of a.

This problem reminds me of a problem from Pre RMO 2017. If you expand the given expression you will get \( a \sqrt a + b \sqrt b - a \sqrt b - b \sqrt a = 1 \). The Pre RMO problem set the values of \( a \sqrt a + b \sqrt b = 183 \) and \( a \sqrt b -+ b \sqrt a = 182 \)