Toolkit 12Sum of cubes (factored)a3+b3=(a+b)(a2−ab+b2)a^3+b^3=(a+b)(a^2-ab+b^2)a3+b3=(a+b)(a2−ab+b2)ProofUsing the distributive law,(a+b)(a2−ab+b2)=a3−a2b+ab2+a2b−ab2+b3=a3+b3.□\begin{aligned} (a+b)(a^2 - ab + b^2) &= a^3 - a^2b + ab^2 + a^2b - ab^2 + b^3 \\ &= a^3 + b^3. \quad\square \end{aligned}(a+b)(a2−ab+b2)=a3−a2b+ab2+a2b−ab2+b3=a3+b3.□Related ProblemsCoreAMC 10A 2020 (Problem 14)