AMC 12A 2025 (Problem 19)Let aaa, bbb, and ccc be the roots of the polynomial x3+kx+1x^3+kx+1x3+kx+1. What is the sum a3b2+a2b3+b3c2+b2c3+c3a2+c2a3?a^3b^2+a^2b^3+b^3c^2+b^2c^3+c^3a^2+c^2a^3?a3b2+a2b3+b3c2+b2c3+c3a2+c2a3?(A) -k\text{(A)}\;\text{-k}(A)-k(B) -k+1\text{(B)}\;\text{-k+1}(B)-k+1(C) 1\text{(C)}\;1(C)1(D) k-1\text{(D)}\;\text{k-1}(D)k-1(E) k\text{(E)}\;\text{k}(E)kRelated TopicsToolkit 30 — Vieta's FormulaHints (7)Hint 1a+b+c=0a+b+c=0a+b+c=0Hint 2ab+ac+bc=kab+ac+bc=kab+ac+bc=kHint 3abc=−1abc=-1abc=−1Hint 4a3b2+a2b3+b3c2+b2c3+c3a2+c2a3=a2b2(a+b)+b2c2(b+c)+c2a2(c+a)a^3b^2+a^2b^3+b^3c^2+b^2c^3+c^3a^2+c^2a^3 = a^2b^2(a+b)+b^2c^2(b+c)+c^2a^2(c+a)a3b2+a2b3+b3c2+b2c3+c3a2+c2a3=a2b2(a+b)+b2c2(b+c)+c2a2(c+a)Hint 5a+b=−c,b+c=−a,c+a=−ba+b=-c,\qquad b+c=-a,\qquad c+a=-ba+b=−c,b+c=−a,c+a=−bHint 6a2b2(a+b)+b2c2(b+c)+c2a2(c+a)=a2b2(−c)+b2c2(−a)+c2a2(−b)a^2b^2(a+b)+b^2c^2(b+c)+c^2a^2(c+a) = a^2b^2(-c)+b^2c^2(-a)+c^2a^2(-b)a2b2(a+b)+b2c2(b+c)+c2a2(c+a)=a2b2(−c)+b2c2(−a)+c2a2(−b)a2b2(−c)+b2c2(−a)+c2a2(−b)=−abc(ab+bc+ca)a^2b^2(-c)+b^2c^2(-a)+c^2a^2(-b) = -abc(ab+bc+ca)a2b2(−c)+b2c2(−a)+c2a2(−b)=−abc(ab+bc+ca)Hint 7−abc(ab+bc+ca)=−(−1)(k)-abc(ab+bc+ca) = -(-1)(k)−abc(ab+bc+ca)=−(−1)(k)Final Answer(E) kkk