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2025 AMC 12A Problem 19

Below is the professionally curated solution for Problem 19 of the 2025 AMC 12A, 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 of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:Vieta’s Formulassymmetry (algebra)

Difficulty rating: 2020

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?a^3b^2 + a^2b^3 + b^3c^2 + b^2c^3 + c^3a^2 + c^2a^3?

−k-k

−k+1-k + 1

11

k−1k - 1

kk

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).a^2b^2(a + b) + b^2c^2(b + c) + c^2a^2(c + a). 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.-a^2b^2 c - ab^2c^2 - a^2bc^2 = -abc(ab + bc + ca) = -(-1)(k) = k.

Thus, the correct answer is E.

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