# 2000 JBMO Problems/Problem 1

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## Problem

Let $x$ and $y$ be positive reals such that $$x^3 + y^3 + (x + y)^3 + 30xy = 2000.$$ Show that $x + y = 10$.

## Solution

Rearranging the equation yields $$x^3 + y^3 + (x + y)^3 + 30xy - 2000 = 0.$$ If $x+y=10$ in the large equation, then $x+y-10$ must be a factor of the large equation. Note that we can rewrite the large equation as \begin{align*} 0 &= (x+y)^3 - 1000 + x^3 + 3x^2y - 3x^2y + 3xy^2 - 3xy^2 + y^3 - 1000 + 30xy \\ &= 2[(x+y)^3 - 1000] - 3x^2y - 3xy^2 + 30xy. \end{align*} We can factor the difference of cubes in the first part and factor $3xy$ in the second part, resulting in $$0 = 2(x+y-10)((x+y)^2 + 10x + 10y + 100) - 3xy(x+y-10)$$ Finally, we can factor by grouping, which results in \begin{align*} 0 &= (x+y-10)(2(x+y)^2 + 20x + 20y + 200 - 3xy) \\ &= (x+y-10)(2x^2 + xy + 2y^2 + 20x + 20y + 200). \end{align*} By the Zero Product Property, either $x+y=10$ or $2x^2 + xy + 2y^2 + 20x + 20y + 200 = 0.$ However, since $x$ and $y$ are both positive, $2x^2 + xy + 2y^2 + 20x + 20y + 200$ can not equal zero, so we have proved that $x+y = 10.$

## See Also

 2000 JBMO (Problems • Resources) Preceded byFirst Problem Followed byProblem 2 1 • 2 • 3 • 4 All JBMO Problems and Solutions
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