2025 AMC 12A Problem 19Problem 19 of 25·Harder·AlgebraLet a,a,a, b,b,b, and ccc be the roots of the polynomial x3+kx+1.x^3 + kx + 1.x3+kx+1. What is the sum a3b2+a2b3+b3c2+b2c3+c3a2+c2a3? \begin{aligned} &a^3b^2 + a^2b^3 + b^3c^2 \\ &\quad {}+ b^2c^3 + c^3a^2 + c^2a^3? \end{aligned} a3b2+a2b3+b3c2+b2c3+c3a2+c2a3?Answer choicesA−k-k−k1B−k+1-k + 1−k+12C1113Dk−1k - 1k−14Ekkk5Submit answerStuck? Show hintsShow solutionSolutionBy Vieta’s formulas, a+b+c=0,a + b + c = 0,a+b+c=0, ab+bc+ca=k,ab + bc + ca = k,ab+bc+ca=k, and abc=−1.abc = -1.abc=−1. Group the sum as a2b2(a+b)+b2c2(b+c)+c2a2(c+a). \begin{aligned} &a^2b^2(a + b) + b^2c^2(b + c) \\ &\quad {}+ c^2a^2(c + a). \end{aligned} a2b2(a+b)+b2c2(b+c)+c2a2(c+a). Since a+b+c=0,a + b + c = 0,a+b+c=0, we have a+b=−c,a + b = -c,a+b=−c, b+c=−a,b + c = -a,b+c=−a, c+a=−b.c + a = -b.c+a=−b. So the sum equals −a2b2c−ab2c2−a2bc2=−abc(ab+bc+ca)=−(−1)(k)=k. \begin{gathered} -a^2b^2 c - ab^2c^2 - a^2bc^2 \\ = -abc(ab + bc + ca) \\ = -(-1)(k) = k. \end{gathered} −a2b2c−ab2c2−a2bc2=−abc(ab+bc+ca)=−(−1)(k)=k. Thus, the correct answer is E.