Let’s analyze and solve this problem step-by-step.

Given:

1. Set $A = \{ x \mid x^2 - 1 = 0 \}$
2. Set $B = \{ x \mid x^2 + x - 2 = 0 \}$
3. Set $C = \{ x \mid x^3 - x^2 = 0 \}$

The task requires finding: a. $A \cup (B \cap C)$ b. $(A \cup B) \cap C$ c. $(A - B) \cup C$

Step 1: Solve for the Sets $A$, $B$, and $C$

1. Set $A$: Solve $x^2 - 1 = 0$. $x^2 - 1 = 0 \Rightarrow (x - 1)(x + 1) = 0$ Therefore, $x = 1$ or $x = -1$. So, $A = \{ -1, 1 \}$.

2. Set $B$: Solve $x^2 + x - 2 = 0$. $x^2 + x - 2 = 0 \Rightarrow (x + 2)(x - 1) = 0$ Therefore, $x = -2$ or $x = 1$. So, $B = \{ -2, 1 \}$.

3. Set $C$: Solve $x^3 - x^2 = 0$. $x^2(x - 1) = 0$ This gives $x = 0$ or $x = 1$. So, $C = \{ 0, 1 \}$.

Step 2: Calculate the Required Expressions

Now, let's determine each expression in the question.

Part (a): $A \cup (B \cap C)$

1. Find $B \cap C$: $B \cap C = \{ -2, 1 \} \cap \{ 0, 1 \} = \{ 1 \}$
2. Now, find $A \cup (B \cap C)$: $A \cup (B \cap C) = \{ -1, 1 \} \cup \{ 1 \} = \{ -1, 1 \}$

Part (b): $(A \cup B) \cap C$

1. Find $A \cup B$: $A \cup B = \{ -1, 1 \} \cup \{ -2, 1 \} = \{ -2, -1, 1 \}$
2. Now, find $(A \cup B) \cap C$: $(A \cup B) \cap C = \{ -2, -1, 1 \} \cap \{ 0, 1 \} = \{ 1 \}$

Part (c): $(A - B) \cup C$

1. Find $A - B$: $A - B = \{ -1, 1 \} - \{ -2, 1 \} = \{ -1 \}$
2. Now, find $(A - B) \cup C$: $(A - B) \cup C = \{ -1 \} \cup \{ 0, 1 \} = \{ -1, 0, 1 \}$

Summary of Solutions

• (a) $A \cup (B \cap C) = \{ -1, 1 \}$
• (b) $(A \cup B) \cap C = \{ 1 \}$
• (c) $(A - B) \cup C = \{ -1, 0, 1 \}$

Would you like further details on any of these steps or have questions about specific parts?

Related Questions to Expand Understanding

1. How do you determine the union and intersection of two sets in general?
2. What is the difference between the union and difference of sets?
3. How would you visualize these sets on a number line?
4. Can you find the symmetric difference between sets $A$ and $B$?
5. What would change if we added another set $D$ defined by a different equation?

Tip

When working with set operations involving unions and intersections, it helps to solve each set individually first before applying the operations. This approach prevents errors and makes it easier to see how each operation affects the result.