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Difference between revisions of "2015 AIME II Problems/Problem 14"

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==Solution==
 
==Solution==
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The expression we want to find is <math>2(x^3+y^3) + x^3y^3</math>.
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Factor the given equations as <math>x^4y^4(x+y) = 810</math> and <math>x^3y^3(x^3+y^3)=945</math>, respectively. Dividing the latter by the former equation yields <math>\frac{x^2-xy+y^2}{xy} = \frac{945}{810}</math>. Adding 3 to both sides and simplifying yields <math>\frac{(x+y)^2}{xy} = \frac{25}{6}</math>. Solving for <math>x+y</math> and substituting this expression into the first equation yields <math>\frac{5\sqrt{6}}{6}(xy)^{\frac{9}{2}} = 810</math>. Solving for <math>xy</math>, we find that <math>xy = 3\sqrt[3]{2}</math>, so <math>x^3y^3 = 54</math>. Substituting this into the second equation and solving for <math>x^3+y^3</math> yields <math>x^3+y^3=\frac{35}{2}</math>. So, the expression to evaluate is equal to <math>2 \times \frac{35}{2} + 54 = \boxed{89}</math>.

Revision as of 21:52, 26 March 2015

Problem

Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$. Evaluate $2x^3+(xy)^3+2y^3$.

Solution

The expression we want to find is $2(x^3+y^3) + x^3y^3$.

Factor the given equations as $x^4y^4(x+y) = 810$ and $x^3y^3(x^3+y^3)=945$, respectively. Dividing the latter by the former equation yields $\frac{x^2-xy+y^2}{xy} = \frac{945}{810}$. Adding 3 to both sides and simplifying yields $\frac{(x+y)^2}{xy} = \frac{25}{6}$. Solving for $x+y$ and substituting this expression into the first equation yields $\frac{5\sqrt{6}}{6}(xy)^{\frac{9}{2}} = 810$. Solving for $xy$, we find that $xy = 3\sqrt[3]{2}$, so $x^3y^3 = 54$. Substituting this into the second equation and solving for $x^3+y^3$ yields $x^3+y^3=\frac{35}{2}$. So, the expression to evaluate is equal to $2 \times \frac{35}{2} + 54 = \boxed{89}$.

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