if A and B are two sets them
`(A-B)cup (B-A)cup (AcapB)`is equal to

Options

A. `A cup B`
B. `Acap B`
C. `A`
D. B'

Explanation

To solve the problem, we need to simplify the expression \((A - B) \cup (B - A) \cup (A \cap B)\).

### Step-by-Step Solution:

1. **Understanding the Sets**:
- \(A - B\) represents the elements that are in set \(A\) but not in set \(B\).
- \(B - A\) represents the elements that are in set \(B\) but not in set \(A\).
- \(A \cap B\) represents the elements that are common to both sets \(A\) and \(B\).

2. **Visualizing with a Venn Diagram**:
- Draw a Venn diagram with two overlapping circles representing sets \(A\) and \(B\).
- Shade the area representing \(A - B\) (the part of circle \(A\) that does not overlap with circle \(B\)).
- Shade the area representing \(B - A\) (the part of circle \(B\) that does not overlap with circle \(A\)).
- The area representing \(A \cap B\) (the overlapping part of both circles) is also shaded.

3. **Combining the Sets**:
- Now, we can combine the shaded areas:
\[ (A - B) \cup (B - A) \cup (A \cap B) \]
- This means we are taking all the elements that are either in \(A\) but not in \(B\), in \(B\) but not in \(A\), or in both \(A\) and \(B\).

4. **Simplifying the Expression**:
- Notice that the union of \(A - B\) and \(B - A\) gives us all elements that are in either \(A\) or \(B\) but not in both. This can be expressed as:
\[ (A - B) \cup (B - A) = A \cup B - (A \cap B) \]
- Therefore, we can rewrite our original expression as:
\[ (A \cup B - (A \cap B)) \cup (A \cap B) \]

5. **Final Simplification**:
- When we take the union of \(A \cup B - (A \cap B)\) with \(A \cap B\), we essentially include all elements from both sets \(A\) and \(B\):
\[ A \cup B \]

### Conclusion:
Thus, we conclude that:
\[ (A - B) \cup (B - A) \cup (A \cap B) = A \cup B \]