1. The problem is to simplify $ (A+B+C)(A+B+C)(A+B+C) $ using logic gates. 2. In Boolean algebra, $ + $ stands for OR, and $ $ stands for AND. 3. First recognize that $ (A+B+C)(A+B+C)(A+B+C) $ is just $ (A+B+C)^3 $. 4. Since in Boolean algebra $ X \times X = X $ (idempotent law), we have $ (A+B+C)^3 = A+B+C $. 5. Thus, $ (A+B+C)(A+B+C)(A+B+C) = A+B+C $. 6. So the expression simplifies to $ A+B+C $ using logic gates. 7. This means the original circuit with three repeated OR gates can be reduced to a single OR gate of inputs $A, B, C$. Final answer: $ (A+B+C)(A+B+C)(A+B+C) = A+B+C $