2010 AIME I Problems/Problem 9

Problem

Let $(a,b,c)$ be the real solution of the system of equations $x^3 - xyz = 2$ , $y^3 - xyz = 6$ , $z^3 - xyz = 20$ . The greatest possible value of $a^3 + b^3 + c^3$ can be written in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .

Solution

Add the three equations to get $a^3 + b^3 + c^3 = 28 + 3abc$ . Now, let $abc = p$ . $a = \sqrt [3]{p + 2}$ , $b = \sqrt [3]{p + 6}$ and $c = \sqrt [3]{p + 20}$ , so $p = abc = (\sqrt [3]{p + 2})(\sqrt [3]{p + 6})(\sqrt [3]{p + 20})$ . Now cube both sides; the $p^3$ terms cancel out. Solve the remaining quadratic to get $p = - 4, - \frac {15}{7}$ . To maximize $a^3 + b^3 + c^3$ choose $p = - \frac {15}{7}$ and so the sum is $28 - \frac {45}{7} = \frac {196 - 45}{7}$ giving $151 + 7 = \fbox{158}$ .

2010 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2010 AIME I Problems/Problem 9

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