------
## 最终结论
To solve the problem of finding the smallest possible value of \((a \oplus x) + (b \oplus x)\) for given integers \(a\) and \(b\), we can utilize the properties of the XOR operation.

The key observation is that to minimize \((a \oplus x) + (b \oplus x)\), we should choose \(x = a \oplus b\). This is because:

- The expression simplifies to \((a \oplus (a \oplus b)) + (b \oplus (a \oplus b))\).
- Simplifying further using properties of XOR, we get: 
  - \(a \oplus (a \oplus b) = b\)
  - \(b \oplus (a \oplus b) = a\)
- Therefore, the expression becomes \(b + a = a + b\).

Thus, the minimum value for each test case is simply \(a + b - 2(a \& b)\), which simplifies to just \(a + b\) when using the optimal choice of \(x = a \oplus b\).

Here's the Python code implementing this logic:

```python
def smallest_xor_sum(t, test_cases):
    results = []
    for a, b in test_cases:
        # Calculate the smallest sum by choosing x = a ^ b
        results.append((a ^ b))
    return results

# Input reading
import sys
input = sys.stdin.read
data = input().split()

t = int(data[0])
test_cases = [(int(data[i * 2 + 1]), int(data[i * 2 + 2])) for i in range(t)]

# Output the results
for result in smallest_xor_sum(t, test_cases):
    print(result)
```

In this code:
- We define a function `smallest_xor_sum` that computes the smallest possible value for each test case.
- We read input values and parse them into test cases.
- For each test case, we compute and print the result.