If a:b=c:d=e:f then shows that (a⁴+a²e²-e⁴)/(b⁶+b²f²−f⁵) = a⁴/b⁴

$$A \cup B = \{1,2,5,6,7\} \cup \{1,3,4,5,6,8\}$$ $$A \cup B = \{1,2,3,4,5,6,7,8\}$$

This includes all elements in the universal set $U$ that are not in $A \cup B$.

$$U = \{1,2,3,4,5,6,7,8,9\}$$ $$(A \cup B)' = U - (A \cup B)$$ $$(A \cup B)' = \{1,2,3,4,5,6,7,8,9\} - \{1,2,3,4,5,6,7,8\}$$ $$(A \cup B)' = \{9\}$$

This includes all elements in $U$ that are not in $A$.

$$A' = U - A$$ $$A' = \{1,2,3,4,5,6,7,8,9\} - \{1,2,5,6,7\}$$ $$A' = \{3,4,8,9\}$$

This includes all elements in $U$ that are not in $B$.

$$B' = U - B$$ $$B' = \{1,2,3,4,5,6,7,8,9\} - \{1,3,4,5,6,8\}$$ $$B' = \{2,7,9\}$$

This includes all elements common to both $A'$ and $B'$.

$$A' \cap B' = \{3,4,8,9\} \cap \{2,7,9\}$$ $$A' \cap B' = \{9\}$$

We found $(A \cup B)' = \{9\}$ and $A' \cap B' = \{9\}$.
Since both sides are equal, the statement is proven.

$$(A \cup B)' = A' \cap B'$$