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

Revision as of 14:26, 2 February 2022

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: $2(5\cdot2^{-1/3})^3-6(15)+54=125-90+54=\boxed{089}$

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 {{AIME box|year=2015|n=II|num-b=13|num-a=15}}
 {{MAA Notice}}
+[[Category:Intermediate Algebra Problems]]

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