2024 AMC 8 Problems/Problem 22

Problem 22

What is the sum of the cubes of the solutions cubed of $x^5+2x^4+3x^3+3x^2+2x+1=0$ ?

(A) 1 (B) 8 (C) 27 (D) -1 (E) -27

Solution

Factoring $x^5+2x^4+3x^3+3x^2+2x+1$ yields $(x+1)(x^2+1)(x^2+x+1)$ . Denote $a, b, c, d, e$ to be solutions of this polynomial. We can easily find one of the solutions is $a=-1$ . Using the quadratic formula on the rest of the factors yields $b=-i, c=i, d=\frac{-1-i\sqrt{3}}{2},$ and finally $e=\frac{-1+i\sqrt{3}}{2}$ . The sum $a^3+b^3+c^3+d^3+e^3$ is 1, so 1 to the third power is 1. So, the final answer is $\boxed{A}$ .

-Rainier2020

Sidenote: You also could have used Newtonian sums to solve this problem.

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2024 AMC 8 Problems/Problem 22

Problem 22

Solution