answer:First, let's find A △ B (the symmetric difference between sets A and B). The symmetric difference between two sets is the set of elements which are in either of the sets but not in their intersection. A △ B = {1, 3, 5, 7, 2, 4, 6, 8} - { } = {1, 3, 5, 7, 2, 4, 6, 8} Now let's find (A △ B) △ {2, 4}: (A △ B) △ {2, 4} = {1, 3, 5, 7, 2, 4, 6, 8} △ {2, 4} The symmetric difference between these two sets is: {1, 3, 5, 7, 6, 8} (since 2 and 4 are removed as they are present in both sets) Now let's find B △ {2, 4}: B △ {2, 4} = {2, 4, 6, 8} △ {2, 4} = {6, 8} (since 2 and 4 are removed as they are present in both sets) Now let's find A △ (B △ {2, 4}): A △ {6, 8} = {1, 3, 5, 7} △ {6, 8} The symmetric difference between these two sets is: {1, 3, 5, 7, 6, 8} We can see that (A △ B) △ {2, 4} = {1, 3, 5, 7, 6, 8} and A △ (B △ {2, 4}) = {1, 3, 5, 7, 6, 8}. Therefore, the two expressions are equal.

answer:First, we need to find the symmetric difference (A Δ B). The symmetric difference is the set of elements which are in either of the sets A or B, but not in their intersection. A Δ B = (A ∪ B) - (A ∩ B) A ∪ B = {1, 2, 3, 4} (union of A and B) A ∩ B = {2, 3} (intersection of A and B) A Δ B = {1, 2, 3, 4} - {2, 3} = {1, 4} Now, we need to find the union of (A Δ B) and B. (A Δ B) ∪ B = {1, 4} ∪ {2, 3, 4} = {1, 2, 3, 4} So, (A Δ B) ∪ B = {1, 2, 3, 4}.

answer:First, we need to find the symmetric difference of A and B (A Δ B). The symmetric difference of two sets is the set of elements which are in either of the sets but not in their intersection. A Δ B = (A ∪ B) - (A ∩ B) A ∪ B = {1, 2, 3, 4, 5, 6, 7} (union of A and B) A ∩ B = {3, 4, 5} (intersection of A and B) Now, subtract the intersection from the union: A Δ B = {1, 2, 6, 7} Next, we need to find the union of (A Δ B) and A: (A Δ B) ∪ A = {1, 2, 6, 7} ∪ {1, 2, 3, 4, 5} The union of these two sets is: {1, 2, 3, 4, 5, 6, 7} So, (A Δ B) ∪ A = {1, 2, 3, 4, 5, 6, 7}.

answer:First, we find the symmetric difference of A and B. The symmetric difference is the set of elements that are in either A or B, but not in both. A ∆ B = {1, 3, 5} ∪ {2, 4, 6} = {1, 2, 3, 4, 5, 6} Now, we find the union of this result with C: (A ∆ B) ∪ C = {1, 2, 3, 4, 5, 6} ∪ {2, 5, 6, 8} = {1, 2, 3, 4, 5, 6, 8} Finally, we find the symmetric difference of the union we just found and A: ({1, 2, 3, 4, 5, 6, 8} ∆ A) = {2, 4, 6, 8} ∪ {1, 3, 7} = {1, 2, 3, 4, 6, 7, 8} So, the symmetric difference of the union and A is {1, 2, 3, 4, 6, 7, 8}.