1. **State the problem:** We want to simplify the expression $$(A+B+C)(A+B+C)(A+B+C)$$ using algebra. 2. **Formula and rules:** This expression is the cube of a trinomial, written as $$(A+B+C)^3$$. 3. **Step 1: Expand the first two factors:** $$(A+B+C)(A+B+C) = A^2 + 2AB + 2AC + B^2 + 2BC + C^2$$ More precisely, using distributive property: $$= A^2 + AB + AC + BA + B^2 + BC + CA + CB + C^2$$ Since multiplication is commutative, $AB = BA$, $AC = CA$, $BC = CB$, so: $$= A^2 + 2AB + 2AC + B^2 + 2BC + C^2$$ 4. **Step 2: Multiply the result by the third factor $(A+B+C)$:** $$ (A^2 + 2AB + 2AC + B^2 + 2BC + C^2)(A+B+C) $$ 5. **Step 3: Distribute each term:** $$= A^2(A+B+C) + 2AB(A+B+C) + 2AC(A+B+C) + B^2(A+B+C) + 2BC(A+B+C) + C^2(A+B+C)$$ 6. **Step 4: Multiply out each term:** $$= A^3 + A^2B + A^2C + 2A^2B + 2AB^2 + 2ABC + 2A^2C + 2ABC + 2AC^2 + AB^2 + B^3 + B^2C + 2ABC + 2B^2C + 2BC^2 + AC^2 + BC^2 + C^3$$ 7. **Step 5: Combine like terms:** $$= A^3 + B^3 + C^3 + 3A^2B + 3AB^2 + 3A^2C + 3AC^2 + 3B^2C + 3BC^2 + 6ABC$$ 8. **Final answer:** $$ (A+B+C)^3 = A^3 + B^3 + C^3 + 3A^2B + 3AB^2 + 3A^2C + 3AC^2 + 3B^2C + 3BC^2 + 6ABC $$ This formula shows how to expand the cube of a sum of three terms by applying distributive property and combining like terms.