Answer :
Let's go through the solution step-by-step.
Given:
1. [tex]\( A = \{ x : x \text{ is a positive integer } < 8 \} \)[/tex], which means [tex]\( A = \{1, 2, 3, 4, 5, 6, 7\} \)[/tex].
2. [tex]\( B = \{ x : x^3 - 6x^2 + 11x - 6 = 0 \} \)[/tex].
First, we need to solve the polynomial equation [tex]\( x^3 - 6x^2 + 11x - 6 = 0 \)[/tex]. This can be factored as follows:
[tex]\[ x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3) \][/tex]
Thus, the roots are [tex]\( x = 1, 2, 3 \)[/tex].
Hence, [tex]\( B = \{1, 2, 3\} \)[/tex].
3. [tex]\( C = \{ x : x \text{ is an even number } < 8 \} \)[/tex].
So, [tex]\( C = \{2, 4, 6\} \)[/tex].
Now, let's verify each part of the question.
### Part a: [tex]\( A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \)[/tex]
1. Determine [tex]\( B \cup C \)[/tex]:
[tex]\[ B \cup C = \{1, 2, 3\} \cup \{2, 4, 6\} = \{1, 2, 3, 4, 6\} \][/tex]
2. Find [tex]\( A \cap (B \cup C) \)[/tex]:
[tex]\[ A \cap (B \cup C) = \{1, 2, 3, 4, 5, 6, 7\} \cap \{1, 2, 3, 4, 6\} = \{1, 2, 3, 4, 6\} \][/tex]
3. Determine [tex]\( A \cap B \)[/tex]:
[tex]\[ A \cap B = \{1, 2, 3, 4, 5, 6, 7\} \cap \{1, 2, 3\} = \{1, 2, 3\} \][/tex]
4. Determine [tex]\( A \cap C \)[/tex]:
[tex]\[ A \cap C = \{1, 2, 3, 4, 5, 6, 7\} \cap \{2, 4, 6\} = \{2, 4, 6\} \][/tex]
5. Find [tex]\( (A \cap B) \cup (A \cap C) \)[/tex]:
[tex]\[ (A \cap B) \cup (A \cap C) = \{1, 2, 3\} \cup \{2, 4, 6\} = \{1, 2, 3, 4, 6\} \][/tex]
Since both sides of the equation match, we have shown that:
[tex]\[ A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \][/tex]
### Part b: [tex]\( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \)[/tex]
1. Determine [tex]\( B \cap C \)[/tex]:
[tex]\[ B \cap C = \{1, 2, 3\} \cap \{2, 4, 6\} = \{2\} \][/tex]
2. Find [tex]\( A \cup (B \cap C) \)[/tex]:
[tex]\[ A \cup (B \cap C) = \{1, 2, 3, 4, 5, 6, 7\} \cup \{2\} = \{1, 2, 3, 4, 5, 6, 7\} \][/tex]
3. Determine [tex]\( A \cup B \)[/tex]:
[tex]\[ A \cup B = \{1, 2, 3, 4, 5, 6, 7\} \cup \{1, 2, 3\} = \{1, 2, 3, 4, 5, 6, 7\} \][/tex]
4. Determine [tex]\( A \cup C \)[/tex]:
[tex]\[ A \cup C = \{1, 2, 3, 4, 5, 6, 7\} \cup \{2, 4, 6\} = \{1, 2, 3, 4, 5, 6, 7\} \][/tex]
5. Find [tex]\( (A \cup B) \cap (A \cup C) \)[/tex]:
[tex]\[ (A \cup B) \cap (A \cup C) = \{1, 2, 3, 4, 5, 6, 7\} \cap \{1, 2, 3, 4, 5, 6, 7\} = \{1, 2, 3, 4, 5, 6, 7\} \][/tex]
Since both sides of the equation match, we have shown that:
[tex]\[ A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \][/tex]
### Part c: [tex]\( A \cap (B \cap C) = (A \cap B) \cap C \)[/tex]
1. We've already determined [tex]\( B \cap C = \{2\} \)[/tex].
2. Find [tex]\( A \cap (B \cap C) \)[/tex]:
[tex]\[ A \cap (B \cap C) = \{1, 2, 3, 4, 5, 6, 7\} \cap \{2\} = \{2\} \][/tex]
3. We've already determined [tex]\( A \cap B = \{1, 2, 3\} \)[/tex].
4. Determine [tex]\( (A \cap B) \cap C \)[/tex]:
[tex]\[ (A \cap B) \cap C = \{1, 2, 3\} \cap \{2, 4, 6\} = \{2\} \][/tex]
Since both sides of the equation match, we have shown that:
[tex]\[ A \cap (B \cap C) = (A \cap B) \cap C \][/tex]
Given:
1. [tex]\( A = \{ x : x \text{ is a positive integer } < 8 \} \)[/tex], which means [tex]\( A = \{1, 2, 3, 4, 5, 6, 7\} \)[/tex].
2. [tex]\( B = \{ x : x^3 - 6x^2 + 11x - 6 = 0 \} \)[/tex].
First, we need to solve the polynomial equation [tex]\( x^3 - 6x^2 + 11x - 6 = 0 \)[/tex]. This can be factored as follows:
[tex]\[ x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3) \][/tex]
Thus, the roots are [tex]\( x = 1, 2, 3 \)[/tex].
Hence, [tex]\( B = \{1, 2, 3\} \)[/tex].
3. [tex]\( C = \{ x : x \text{ is an even number } < 8 \} \)[/tex].
So, [tex]\( C = \{2, 4, 6\} \)[/tex].
Now, let's verify each part of the question.
### Part a: [tex]\( A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \)[/tex]
1. Determine [tex]\( B \cup C \)[/tex]:
[tex]\[ B \cup C = \{1, 2, 3\} \cup \{2, 4, 6\} = \{1, 2, 3, 4, 6\} \][/tex]
2. Find [tex]\( A \cap (B \cup C) \)[/tex]:
[tex]\[ A \cap (B \cup C) = \{1, 2, 3, 4, 5, 6, 7\} \cap \{1, 2, 3, 4, 6\} = \{1, 2, 3, 4, 6\} \][/tex]
3. Determine [tex]\( A \cap B \)[/tex]:
[tex]\[ A \cap B = \{1, 2, 3, 4, 5, 6, 7\} \cap \{1, 2, 3\} = \{1, 2, 3\} \][/tex]
4. Determine [tex]\( A \cap C \)[/tex]:
[tex]\[ A \cap C = \{1, 2, 3, 4, 5, 6, 7\} \cap \{2, 4, 6\} = \{2, 4, 6\} \][/tex]
5. Find [tex]\( (A \cap B) \cup (A \cap C) \)[/tex]:
[tex]\[ (A \cap B) \cup (A \cap C) = \{1, 2, 3\} \cup \{2, 4, 6\} = \{1, 2, 3, 4, 6\} \][/tex]
Since both sides of the equation match, we have shown that:
[tex]\[ A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \][/tex]
### Part b: [tex]\( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \)[/tex]
1. Determine [tex]\( B \cap C \)[/tex]:
[tex]\[ B \cap C = \{1, 2, 3\} \cap \{2, 4, 6\} = \{2\} \][/tex]
2. Find [tex]\( A \cup (B \cap C) \)[/tex]:
[tex]\[ A \cup (B \cap C) = \{1, 2, 3, 4, 5, 6, 7\} \cup \{2\} = \{1, 2, 3, 4, 5, 6, 7\} \][/tex]
3. Determine [tex]\( A \cup B \)[/tex]:
[tex]\[ A \cup B = \{1, 2, 3, 4, 5, 6, 7\} \cup \{1, 2, 3\} = \{1, 2, 3, 4, 5, 6, 7\} \][/tex]
4. Determine [tex]\( A \cup C \)[/tex]:
[tex]\[ A \cup C = \{1, 2, 3, 4, 5, 6, 7\} \cup \{2, 4, 6\} = \{1, 2, 3, 4, 5, 6, 7\} \][/tex]
5. Find [tex]\( (A \cup B) \cap (A \cup C) \)[/tex]:
[tex]\[ (A \cup B) \cap (A \cup C) = \{1, 2, 3, 4, 5, 6, 7\} \cap \{1, 2, 3, 4, 5, 6, 7\} = \{1, 2, 3, 4, 5, 6, 7\} \][/tex]
Since both sides of the equation match, we have shown that:
[tex]\[ A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \][/tex]
### Part c: [tex]\( A \cap (B \cap C) = (A \cap B) \cap C \)[/tex]
1. We've already determined [tex]\( B \cap C = \{2\} \)[/tex].
2. Find [tex]\( A \cap (B \cap C) \)[/tex]:
[tex]\[ A \cap (B \cap C) = \{1, 2, 3, 4, 5, 6, 7\} \cap \{2\} = \{2\} \][/tex]
3. We've already determined [tex]\( A \cap B = \{1, 2, 3\} \)[/tex].
4. Determine [tex]\( (A \cap B) \cap C \)[/tex]:
[tex]\[ (A \cap B) \cap C = \{1, 2, 3\} \cap \{2, 4, 6\} = \{2\} \][/tex]
Since both sides of the equation match, we have shown that:
[tex]\[ A \cap (B \cap C) = (A \cap B) \cap C \][/tex]
