2025 AMC 12A Problems/Problem 19

Problem

Let $a$ , $b$ , and $c$ be the roots of the polynomial $x^3 + kx + 1$ . What is the sum $a^3b^2 + a^2b^3 + b^3c^2 + b^2c^3 + c^3a^2 + c^2a^3?$ $\textbf{(A)}~-k\qquad\textbf{(B)}~-k+1\qquad\textbf{(C)}~1\qquad\textbf{(D)}~k-1\qquad\textbf{(E)}~k$

Solution 1

We begin by factoring: \begin{equation} \begin{split} &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).\\ \end{split} \end{equation} From Vieta's Formulas, we know that $a+b+c = 0$ , $ab+bc+ca = k$ , and $abc = -1$ . Therefore, the answer equals $(a^2b^2+b^2c^2+c^2a^2)(0) - (-1)(k) = \boxed{k}.$

~lprado

Solution 2

Notice that this is $\sum_{\text{cyc}} a^2b^2(a+b)$ . By Vieta's (just like in solution 1), we have $a+b+c=0$ , so $a+b=-c$ , $abc=-1$ , and $ab+bc+ac=k$ , so this is equivalent to $\sum_{\text{cyc}} -a^2b^2c = \sum_{\text{cyc}} -(abc)(ab) = \sum_{\text{cyc}} ab = k$ by Vieta's. ~ScoutViolet

Solution 3 (BS)

Let’s just pick a nice and convenient a, b, c. Observe that $2\cdot2^{-\frac13}-2^\frac23=0$ , so we can just have $(a,b,c)=(2^{-\frac13}, 2^{-\frac13},-2^\frac23$ ). So expanding $(x-a)(x-b)(x-c)$ we get $x^3-3\cdot2^{-\frac23}+1$ . Evaluating $a^3b^2 + a^2b^3 + b^3c^2 + b^2c^3 + c^3a^2 + c^2a^3$ , or $2a^5+2a^2c^3+2a^3c^2$ , we get $-3\cdot2^{-\frac23}=k$ , so the answer is $\boxed{\textit{E}}$ . (i did this on the actual test)

~benjamintontungtungtungsahur

who is going to think of this on the test bro 😭😭 and change ur username dawg 🥀🥀

Solution 4 (Vieta's)

First, note that $a+b+c=0$ , $\sum_{cyc} ab = k$ , $abc=-1$ . We can factor the original expression into:

$\sum_{cyc} a^2b^2(a+b) = \sum_{cyc} \left(-\frac{1}{c}\right)^2(-c) = -\frac{\sum_{cyc} ab}{abc} = \boxed{\text{(E) } k.}$

~ABC09090927

Solution 5 (Newton's Sums) (If you suck at factoring like me)

Note that $a^3b^2+a^2b^3+b^3c^2+b^2c^3+c^3a^2+c^2a^3 = (a^3+b^3+c^3)(a^2+b^2+c^2)-a^5+b^5+c^5$

By Newton's sums, we have: $(a^3 + b^3 + c^3) + k(a+b+c) + 3 = 0$ $(a^5 + b^5 + c^5) + k(a^3+b^3+c^3) + 1(a^2+b^2+c^2) = 0$

Note that $a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca)$

From Vieta's formulas, we have $a+b+c = 0$ and $ab + bc + ca = k$

Solving, we get: $(a^3 + b^3 + c^3) = -3$ $(a^2 + b^2 + c^2) = -2k$ $(a^5 + b^5 + c^5) - 3k - 2k = 0 \to (a^5 + b^5 + c^5) = 5k$

Therefore, $a^3b^2+a^2b^3+b^3c^2+b^2c^3+c^3a^2+c^2a^3 = (-3)(-2k)-5k = \boxed{\textbf{(E) }k}$

~ku220

Solution 6

Llike in solution 1, we begin by factoring: \begin{equation} \begin{split} &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)\\ \end{split} \end{equation} From Vieta's Formulas, we know that $a+b+c = 0$ , $ab+bc+ca = k$ , and $abc = -1$ .

square $a+b+c = 0$ to get $a^2 + b^2 + c^2 = 0$ , then isolating one variable each time yields:

\begin{equation} \begin{split} &=-\frac{1}{c^2}(a+b)-\frac{1}{a^2}(b+c)-\frac{1}{b^2}(c+a)\\ \end{split} \end{equation}

Isolating a + b + c = 0, you get

\begin{equation} \begin{split} &=-\frac{1}{c^2}(-c)-\frac{1}{a^2}(-a)-\frac{1}{b^2}(-b)\\ &=-(\frac{ab+bc+ac}{abc})\\ \end{split} \end{equation}

This simplifies to:

\begin{equation} \begin{split} &=-(\frac{k}{-1})\\ &=k \end{split} \end{equation}

Therefore, the answer equals = $\boxed{k}.$

~BOTNATE

Solution 7 (Vieta + Factorization)

We can easily factor the expression into $a^2b^2(a+b)+b^2c^2(b+c)+a^2c^2(a+c)$

By Vieta, we get $a+b+c=0$ , so we can rewrite the expression into $a^2b^2(-c)+b^2c^2(-a)+a^2c^2(-b)$ which is just $-abc(ab+bc+ac)$ . By Vieta we also get $abc=-1$ and $ab+bc+ac=k$ . Therefore the expression is just $-(-1)k=$ $\boxed{k}.$

~backtosq-1

Video Solution 1 by OmegaLearn

https://youtu.be/FuJbXC8x-R

Video Solution 2 by StressedPineapple

https://youtube.com/watch?v=NWBPm3lThH4&t=319s

Video Solution 3 by SpreadTheMathLove

https://www.youtube.com/watch?v=xipj4q--LCk

2025 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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