Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.0 AIME Find the answer

Let $f(x)=x^{3}+3 x-1$ 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 $100 m+n$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

We know that $a^{3}=-3 a+1$ and similarly for $b, c$ , so $\frac{1}{a^{3}+b^{3}}=\frac{1}{2-3 a-3 b}=\frac{1}{2+3 c}=\frac{1}{3(2 / 3+c)}$ Now, $f(x-2 / 3)=x^{3}-2 x^{2}+\frac{13}{3} x-\frac{89}{27}$ 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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