Certainly! Let's begin by defining the sets given in the problem and then showing the required set relationship. Here are the sets:
- [tex]\( A \)[/tex]: [tex]\( A = \{ x: x \text{ is a positive integer } < 8 \} \)[/tex]. This means [tex]\( A = \{ 1, 2, 3, 4, 5, 6, 7 \} \)[/tex].
- [tex]\( B \)[/tex]: [tex]\( B = \left\{ x : x^3 - 6x^2 + 11x - 6 = 0 \right\} \)[/tex]. Solving this equation, we find the roots, which are the integers [tex]\( B = \{ 1, 2, 3 \} \)[/tex].
- [tex]\( C \)[/tex]: [tex]\( C = \{ x : x \text{ is even and } x < 8 \} \)[/tex]. Thus, [tex]\( C = \{ 0, 2, 4, 6 \} \)[/tex].
Next, we follow the steps to solve the expression [tex]\( A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \)[/tex].
### Step-by-Step Solution:
1. Find the Union of [tex]\( B \)[/tex] and [tex]\( C \)[/tex]:
[tex]\[
B \cup C = \{ 1, 2, 3 \} \cup \{ 0, 2, 4, 6 \} = \{ 0, 1, 2, 3, 4, 6 \}
\][/tex]
2. Find the Intersection of [tex]\( A \)[/tex] with [tex]\( B \cup C \)[/tex]:
[tex]\[
A \cap (B \cup C) = \{ 1, 2, 3, 4, 5, 6, 7 \} \cap \{ 0, 1, 2, 3, 4, 6 \} = \{ 1, 2, 3, 4, 6 \}
\][/tex]
3. Find the Intersection of [tex]\( A \)[/tex] with [tex]\( B \)[/tex]:
[tex]\[
A \cap B = \{ 1, 2, 3, 4, 5, 6, 7 \} \cap \{ 1, 2, 3 \} = \{ 1, 2, 3 \}
\][/tex]
4. Find the Intersection of [tex]\( A \)[/tex] with [tex]\( C \)[/tex]:
[tex]\[
A \cap C = \{ 1, 2, 3, 4, 5, 6, 7 \} \cap \{ 0, 2, 4, 6 \} = \{ 2, 4, 6 \}
\][/tex]
5. Find the Union of [tex]\( A \cap B \)[/tex] and [tex]\( A \cap C \)[/tex]:
[tex]\[
(A \cap B) \cup (A \cap C) = \{ 1, 2, 3 \} \cup \{ 2, 4, 6 \} = \{ 1, 2, 3, 4, 6 \}
\][/tex]
### Conclusion:
From the above steps, we find that:
[tex]\[
A \cap (B \cup C) = \{ 1, 2, 3, 4, 6 \}
\][/tex]
and
[tex]\[
(A \cap B) \cup (A \cap C) = \{ 1, 2, 3, 4, 6 \}
\][/tex]
Since both sides of the equation yield the same set, we have shown that:
[tex]\[
A \cap (B \cup C) = (A \cap B) \cup (A \cap C)
\][/tex]
Thus, the equality is proved.
