Toolkit 8Cube of a sum(a+b)3=a3+3a2b+3ab2+b3(a+b)^3=a^3+3a^2b+3ab^2+b^3(a+b)3=a3+3a2b+3ab2+b3ProofBy Toolkit 6 — Square of a sum,(a+b)2=a2+2ab+b2.(a+b)^2 = a^2 + 2ab + b^2.(a+b)2=a2+2ab+b2.Therefore,(a+b)3=(a+b)(a+b)2=(a+b)(a2+2ab+b2)=a3+2a2b+ab2+a2b+2ab2+b3=a3+3a2b+3ab2+b3.□\begin{aligned} (a+b)^3 &= (a+b)(a+b)^2 \\ &= (a+b)(a^2 + 2ab + b^2) \\ &= a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3 \\ &= a^3 + 3a^2b + 3ab^2 + b^3. \quad\square \end{aligned}(a+b)3=(a+b)(a+b)2=(a+b)(a2+2ab+b2)=a3+2a2b+ab2+a2b+2ab2+b3=a3+3a2b+3ab2+b3.□Related ProblemsMajorAMC 12A 2022 (Problem 14)AMC 12B 2025 (Problem 8)