2025 AMC 12A Problem 19

Attempt Problem 19 of the 2025 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

19.

Let a,a, b,b, and cc be the roots of the polynomial x3+kx+1.x^3 + 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}
−k-k
−k+1-k + 1
11
k−1k - 1
kk
Answer: E
Concepts:Vieta’s Formulassymmetry (algebra)
Small hint:
By Vieta, a+b+c=0,a + b + c = 0, ab+bc+ca=k,ab + bc + ca = k, and abc=−1abc = -1
Big hint:
Group as a2b2(a+b)+b2c2(b+c)a^2b^2(a + b) + b^2c^2(b + c) +c2a2(c+a)+ c^2a^2(c + a) and replace a+b=−ca + b = -c
Solution:
By Vieta’s formulas, a+b+c=0,a + b + c = 0, ab+bc+ca=k,ab + bc + ca = k, and 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} Since a+b+c=0,a + b + c = 0, we have a+b=−c,a + b = -c, b+c=−a,b + c = -a, 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}
Thus, the correct answer is E.
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