# 2015 AIME II Problems/Problem 14

## 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 $2 \times \frac{35}{2} + 54 = \boxed{089}$.

Note that since the value we want to find is $2(x^3+y^3)+x^3y^3$, we can convert $2(x^3+y^3)$ into an expression in terms of $x^3y^3$, since from the second equation which is $x^3y^3(x^3+y^3)=945$, we see that $2(x^3+y^3)=1890+x^6y^y,$ and thus the value is $\frac{1890+x^6y^6}{x^3y^3}.$ Since we've already found $x^3y^3,$ we substitute and find the answer to be 89.

## Solution 2

Factor the given equations as $x^4y^4(x+y) = 810$ and $x^3y^3(x+y)(x^2-xy+y^2)=945$, respectively. By the first equation, $x+y=\frac{810}{x^4y^4}$. Plugging this in to the second equation and simplifying yields $(\frac{x}{y}-1+\frac{y}{x})=\frac{7}{6}$. Now substitute $\frac{x}{y}=a$. Solving the quadratic in $a$, we get $a=\frac{x}{y}=\frac{2}{3}$ or $\frac{3}{2}$ As both of the original equations were symmetric in $x$ and $y$, WLOG, let $\frac{x}{y}=\frac{2}{3}$, so $x=\frac{2}{3}y$. Now plugging this in to either one of the equations, we get the solutions $y=\frac{3(2^{\frac{2}{3}})}{2}$, $x=2^{\frac{2}{3}}$. Now plugging into what we want, we get $8+54+27=\boxed{089}$

## Solution 3

Add three times the first equation to the second equation and factor to get $(xy)^3(x^3+3x^2y+3xy^2+y^3)=(xy)^3(x+y)^3=3375$. Taking the cube root yields $xy(x+y)=15$. Noting that the first equation is $(xy)^3\cdot(xy(x+y))=810$, we find that $(xy)^3=\frac{810}{15}=54$. Plugging this into the second equation and dividing yields $x^3+y^3 = \frac{945}{54} = \frac{35}{2}$. Thus the sum required, as noted in Solution 1, is $54+\frac{35}{2}\cdot2 = \boxed{089}$.

## Solution 4

As with the other solutions, factor. But this time, let $a=xy$ and $b=x+y$. Then $a^4b=810$. Notice that $x^3+y^3 = (x+y)(x^2-xy+y^2) = b(b^2-3a)$. Now, if we divide the second equation by the first one, we get $7/6 = \frac{b^2-3a}{a}$; then $\frac{b^2}{a}=\frac{25}{6}$. Therefore, $a = \frac{6}{25}b^2$. Substituting $a$ into $b$ in equation 2 gives us $b^3 = \frac{5^3}{2}$; we are looking for $2b(b^2-3a)+a^3$. Finding $a$, we get $35$. Substituting into the first equation, we get $b=54$. Our final answer is $35+54=\boxed{089}$.

## Solution 5

Factor the given equations as: $$x^4y^4(x+y)=810$$ $$x^3y^3(x^3+y^3)=x^3y^3(x+y)(x^2-xy+y^2)=945$$ We note that these expressions (as well as the desired expression) can be written exclusively in terms of $x+y$ and $xy$. We make the substitution $s=x+y$ and $p=xy$ (for sum and product, respectively).

$$x^4y^4(x+y)=p^4s=810$$ $$x^3y^3(x+y)(x^2-xy+y^2)=p^3(s)(s^2-3p)=s^3p^3-3p^4s=945$$

We see that $p^4s$ shows up in both equations, so we can eliminate it and find $sp$, after which we can get $p^3$ from the first equation. If you rewrite the desired expression using $s$ and $p$, it becomes clear that you don't need to actually find the values of $s$ and $p$, but I will do so for the sake of completion.

$$s^3p^3=945+3p^4s$$ $$s^3p^3=945+3(810)=3375$$ $$sp=15$$

$$p^3=\frac{810}{sp}=54$$ $$p=3\cdot2^{1/3}$$ $$s=\frac{15}{p}=5\cdot2^{-1/3}$$

The desired expression can be written as: $$2(x^3+y^3)+(xy)^3=2(x+y)(x^2-xy+y^2)+(xy)^3$$ $$2(s)(s^2-3p)+p^3=2s^3-6sp+p^3$$

Plugging in $s$ and $p$, we get: $$2(5\cdot2^{-1/3})^3-6(15)+54=125-90+54=\boxed{089}$$

- gting

## See also

 2015 AIME II (Problems • Answer Key • Resources) Preceded byProblem 13 Followed byProblem 15 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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