Let `a^(2)+b^(2)=alpha^(2)+beta^(2)=2`. Then show that the maximum value of `S=(1-a)(1-b)+(1-alpha)(1-beta)` is 8.

Explanation

To solve the problem, we need to find the maximum value of \( S = (1 - a)(1 - b) + (1 - \alpha)(1 - \beta) \) given the conditions \( a^2 + b^2 = 2 \) and \( \alpha^2 + \beta^2 = 2 \).

### Step-by-Step Solution

1. **Substitution**:
We can express \( a \), \( b \), \( \alpha \), and \( \beta \) in terms of trigonometric functions. Let:
\[
a = \sqrt{2} \cos \theta, \quad b = \sqrt{2} \sin \theta, \quad \alpha = \sqrt{2} \cos \phi, \quad \beta = \sqrt{2} \sin \phi
...