If a:b=c:d=e:f then shows that (a^4+a^2e^2-e^4)/(b^6+b^2f^2-f^5) = a^4/b^4

Question

If a:b=c:d=e:f then shows that (a^4+a^2e^2-e^4)/(b^6+b^2f^2-f^5) = a^4/b^4

Asked on March 26, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-03-26T14:02:43.512Z

Step 1: Find the union of sets $A$ and $B$, $A \cup B$. $$A \cup B = \{1,2,5,6,7\} \cup \{1,3,4,5,6,8\}$$ $$A \cup B = \{1,2,3,4,5,6,7,8\}$$ Step 2: Find the complement of $A \cup B$, denoted as $(A \cup B)'$. This includes all elements in the universal set $U$ that are not in $A \cup B$. $$U = \{1,2,3,4,5,6,7,8,9\}$$ $$(A \cup B)' = U - (A \cup B)$$ $$(A \cup B)' = \{1,2,3,4,5,6,7,8,9\} - \{1,2,3,4,5,6,7,8\}$$ $$(A \cup B)' = \{9\}$$ Step 3: Find the complement of set $A$, denoted as $A'$. This includes all elements in $U$ that are not in $A$. $$A' = U - A$$ $$A' = \{1,2,3,4,5,6,7,8,9\} - \{1,2,5,6,7\}$$ $$A' = \{3,4,8,9\}$$ Step 4: Find the complement of set $B$, denoted as $B'$. This includes all elements in $U$ that are not in $B$. $$B' = U - B$$ $$B' = \{1,2,3,4,5,6,7,8,9\} - \{1,3,4,5,6,8\}$$ $$B' = \{2,7,9\}$$ Step 5: Find the intersection of $A'$ and $B'$, denoted as $A' \cap B'$. This includes all elements common to both $A'$ and $B'$. $$A' \cap B' = \{3,4,8,9\} \cap \{2,7,9\}$$ $$A' \cap B' = \{9\}$$ Step 6: Compare the results from Step 2 and Step 5. We found $(A \cup B)' = \{9\}$ and $A' \cap B' = \{9\}$. Since both sides are equal, the statement is proven. $$\boxed{(A \cup B)' = A' \cap B'}$$

Accepted Answer · 2026-03-26T14:02:43.512Z

Step 1: Find the union of sets $A$ and $B$, $A \cup B$. $$A \cup B = \{1,2,5,6,7\} \cup \{1,3,4,5,6,8\}$$ $$A \cup B = \{1,2,3,4,5,6,7,8\}$$ Step 2: Find the complement of $A \cup B$, denoted as $(A \cup B)'$. This includes all elements in the universal set $U$ that are not in $A \cup B$. $$U = \{1,2,3,4,5,6,7,8,9\}$$ $$(A \cup B)' = U - (A \cup B)$$ $$(A \cup B)' = \{1,2,3,4,5,6,7,8,9\} - \{1,2,3,4,5,6,7,8\}$$ $$(A \cup B)' = \{9\}$$ Step 3: Find the complement of set $A$, denoted as $A'$. This includes all elements in $U$ that are not in $A$. $$A' = U - A$$ $$A' = \{1,2,3,4,5,6,7,8,9\} - \{1,2,5,6,7\}$$ $$A' = \{3,4,8,9\}$$ Step 4: Find the complement of set $B$, denoted as $B'$. This includes all elements in $U$ that are not in $B$. $$B' = U - B$$ $$B' = \{1,2,3,4,5,6,7,8,9\} - \{1,3,4,5,6,8\}$$ $$B' = \{2,7,9\}$$ Step 5: Find the intersection of $A'$ and $B'$, denoted as $A' \cap B'$. This includes all elements common to both $A'$ and $B'$. $$A' \cap B' = \{3,4,8,9\} \cap \{2,7,9\}$$ $$A' \cap B' = \{9\}$$ Step 6: Compare the results from Step 2 and Step 5. We found $(A \cup B)' = \{9\}$ and $A' \cap B' = \{9\}$. Since both sides are equal, the statement is proven. $$\boxed{(A \cup B)' = A' \cap B'}$$