Difference between revisions of "2024 AMC 8 Problems/Problem 22"
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==Solution== | ==Solution== | ||
| − | Factoring <math>x^5+2x^4+3x^3+3x^2+2x+1</math> yields <math>(x+1)(x^2+1)(x^2+x+1)</math>. Denote <math>a, b, c, d, e</math> to be solutions of this polynomial. We can easily find one of the solutions is <math>a=-1</math>. Using the quadratic formula on the rest of the factors yields <math>b=-i, c=i, d=\frac{-1-i\sqrt{3}}{2},</math> and finally <math>e=\frac{-1+i\sqrt{3}}{2}</math>. The sum <math>a^3+b^3+c^3+d^3+e^3</math> is 1, so 1 to the third power is <math>\boxed{ | + | Factoring <math>x^5+2x^4+3x^3+3x^2+2x+1</math> yields <math>(x+1)(x^2+1)(x^2+x+1)</math>. Denote <math>a, b, c, d, e</math> to be solutions of this polynomial. We can easily find one of the solutions is <math>a=-1</math>. Using the quadratic formula on the rest of the factors yields <math>b=-i, c=i, d=\frac{-1-i\sqrt{3}}{2},</math> and finally <math>e=\frac{-1+i\sqrt{3}}{2}</math>. The sum <math>a^3+b^3+c^3+d^3+e^3</math> is 1, so 1 to the third power is 1. So, the final answer is <math>\boxed{A}</math>. |
-Rainier2020 | -Rainier2020 | ||
Sidenote: You also could have used Newtonian sums to solve this problem. | Sidenote: You also could have used Newtonian sums to solve this problem. | ||
Revision as of 15:18, 21 January 2024
Problem 22
What is the sum of the cubes of the solutions cubed of 
?
(A) 1 (B) 8 (C) 27 (D) -1 (E) -27
Solution
Factoring 
yields 
. Denote 
to be solutions of this polynomial. We can easily find one of the solutions is 
. Using the quadratic formula on the rest of the factors yields 
and finally 
. The sum 
is 1, so 1 to the third power is 1. So, the final answer is 
.
-Rainier2020
Sidenote: You also could have used Newtonian sums to solve this problem.
