, 2021

Algebra Difficulty 5.0 AIME, harder Prove it United States

Problem:
Let $f(x) = x^{3} + 3x - 1$ have roots $a, b, c$ . Given that
$\frac{1}{a^{3} + b^{3}} + \frac{1}{b^{3} + c^{3}} + \frac{1}{c^{3} + a^{3}}$
can be written as $\frac{m}{n}$ , where $m, n$ are positive integers and $\operatorname{gcd}(m, n) = 1$ , find $100m + n$ .

Solution

Solution:
We know that $a^{3} = -3a + 1$ and similarly for $b, c$ , so
$\frac{1}{a^{3} + b^{3}} = \frac{1}{2 - 3a - 3b} = \frac{1}{2 + 3c} = \frac{1}{3(2/3 + c)}.$
Now,
$f(x - 2/3) = x^{3} - 2x^{2} + \frac{13}{3}x - \frac{89}{27}$
has roots $a + 2/3$ , $b + 2/3$ , and $c + 2/3$ . Thus the answer is, by Vieta's formulas,
$\frac{1}{3} \frac{(a + 2/3)(b + 2/3) + (a + 2/3)(c + 2/3) + (b + 2/3)(c + 2/3)}{(a + 2/3)(b + 2/3)(c + 2/3)} = \frac{13/3}{3 \cdot 89/27} = \frac{39}{89}$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.