Toolkit 13General difference of powersan−bn=(a−b)(an−1+an−2b+⋯+bn−1)a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2}b+\cdots+b^{n-1}\right)an−bn=(a−b)(an−1+an−2b+⋯+bn−1)ProofExpanding the right-hand side gives(a−b)(an−1+an−2b+⋯+abn−2+bn−1)=(an+an−1b+⋯+a2bn−2+abn−1)−(an−1b+an−2b2+⋯+abn−1+bn).\begin{aligned} &(a-b)\left(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n-1}\right) \\ &= \left(a^n + a^{n-1}b + \cdots + a^2 b^{n-2} + ab^{n-1}\right) \\ &\quad - \left(a^{n-1}b + a^{n-2}b^2 + \cdots + ab^{n-1} + b^n\right). \end{aligned}(a−b)(an−1+an−2b+⋯+abn−2+bn−1)=(an+an−1b+⋯+a2bn−2+abn−1)−(an−1b+an−2b2+⋯+abn−1+bn).All intermediate terms cancel, leavingan−bn.□a^n - b^n. \quad\squarean−bn.□Related ProblemsCoreAMC 10B 2022 (Problem 17)