Let us find the disjunctive normal form of E(x,y,z)=(x∧¬z)∨(y∧z).
E(x,y,z)=(x∧¬z)∨(y∧z)=(x∧T∧¬z)∨(T∧y∧z)=(x∧(y∨¬y)∧¬z)∨((x∨¬x)∧y∧z)=(x∧y∧¬z)∨(x∧¬y∧¬z)∨(x∧y∧z)∨(¬x∧y∧z).
It follows that (x∧y∧¬z)∨(x∧¬y∧¬z)∨(x∧y∧z)∨(¬x∧y∧z) is a disjunctive normal form of E(x,y,z).
Let us find the conjunctive normal form of E(x,y,z)=(x∧¬z)∨(y∧z).
E(x,y,z)=(x∧¬z)∨(y∧z)=(x∨y)∧(x∨z)∧(¬z∨y)∧(¬z∨z)=(x∨y∨F)∧(x∨F∨z)∧(F∨y∨¬z)∧T=(x∨y∨(¬z∧z))∧(x∨(¬y∧y)∨z)∧((¬x∧x)∨y∨¬z)=(x∨y∨¬z)∧(x∨y∨z)∧(x∨¬y∨z)∧(x∨y∨z)∧(¬x∨y∨¬z)∧(x∨y∨¬z)=(x∨y∨¬z)∧(x∨y∨z)∧(x∨¬y∨z)∧(¬x∨y∨¬z).
It follows that (x∨y∨¬z)∧(x∨y∨z)∧(x∨¬y∨z)∧(¬x∨y∨¬z) is a conjunctive normal form of E(x,y,z).