AMC12 2025 A

AMC12 2025 A · Q19

AMC12 2025 A · Q19. It mainly tests Vieta / quadratic relationships (basic), Algebra misc.

Let $a$, $b$, and $c$ be the roots of the polynomial $x^3 + kx + 1$. What is the sum\[a^3b^2 + a^2b^3 + b^3c^2 + b^2c^3 + c^3a^2 + c^2a^3?\]

设 $a$，$b$，$c$ 是多项式 $x^3 + kx + 1$ 的根。求 \[a^3b^2 + a^2b^3 + b^3c^2 + b^2c^3 + c^3a^2 + c^2a^3\] 的值。

(A) -k -k

(B) -k+1 -k+1

(D) k-1 k-1

(E) k k

Answer

Correct choice: (E)

正确答案：（E）

Solution

We begin by factoring: a3b2+a2b3+b3c2+b2c3+c3a2+c2a3=a2b2(a+b)+b2c2(b+c)+c2a2(c+a)=(a2b2+b2c2+c2a2)(a+b+c)−a2b2c−b2c2a−c2a2b=(a2b2+b2c2+c2a2)(a+b+c)−(abc)(ab+bc+ca). From Vieta's Formulas, we know that $a+b+c = 0$, $ab+bc+ca = k$, and $abc = -1$. Therefore, the answer equals $(a^2b^2+b^2c^2+c^2a^2)(0) - (-1)(k) = \boxed{k}.$

首先因式分解： $a^3b^2+a^2b^3+b^3c^2+b^2c^3+c^3a^2+c^2a^3=a^2b^2(a+b)+b^2c^2(b+c)+c^2a^2(c+a)=(a^2b^2+b^2c^2+c^2a^2)(a+b+c)−a^2b^2c−b^2c^2a−c^2a^2b=(a^2b^2+b^2c^2+c^2a^2)(a+b+c)−(abc)(ab+bc+ca)$。由 Vieta 公式知 $a+b+c = 0$，$ab+bc+ca = k$，$abc = -1$。因此答案等于 $(a^2b^2+b^2c^2+c^2a^2)(0) - (-1)(k) = \boxed{k}$。

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