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Difference between revisions of "2015 AIME II Problems/Problem 14"
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Let <math>x^2+y^2=13k; xy=6k</math>, then we can see <math>(x+y)^2-12k=13k, x+y=5\sqrt{k}</math>, now, we see <math>x^4y^4\cdot (x+y)=1296k^4\cdot 5\sqrt{k}, k=\frac{1}{\sqrt[3]{4}}</math>.
 
Let <math>x^2+y^2=13k; xy=6k</math>, then we can see <math>(x+y)^2-12k=13k, x+y=5\sqrt{k}</math>, now, we see <math>x^4y^4\cdot (x+y)=1296k^4\cdot 5\sqrt{k}, k=\frac{1}{\sqrt[3]{4}}</math>.
  
−
The rest is easy, <math>2x^3+y^3+x^3y^3=216k^3+2[(x+y)^3-3xy(x+y)]=216k^3+2\cdot 35k^{\frac{3}{2}}=\boxed{89}</math>
+
The rest is easy, <math>2(x^3+y^3)+x^3y^3=216k^3+2[(x+y)^3-3xy(x+y)]=216k^3+2\cdot 35k^{\frac{3}{2}}=\boxed{89}</math>
  
 
~bluesoul
 
~bluesoul
 +
 
==See also==
 
==See also==
 
{{AIME box|year=2015|n=II|num-b=13|num-a=15}}
 
{{AIME box|year=2015|n=II|num-b=13|num-a=15}}

Revision as of 15:22, 28 July 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 $x^3+y^3 = (x+y)(x^2-xy+y^2) = b(b^2-3a)$, so we are looking for $2b(b^2-3a)+a^3$. 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 $\frac{6}{25}b^2$ for $a$ in equation 1, simplifying, and then taking the cube root gives us $b^3 = \frac{5^3}{2}.$ Finding $a^3$ by cubing $a = \frac{6}{25}b^2$ on both sides and simplifying using our previous substitution, we get $a^3 = 54$. Substituting this into the first equation and then dividing by $27$, we get $2b(b^2 - 3a) = 35$. 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

Solution 6

Factor the first and second equations as $(xy)^4(x+y)=810$ and $(xy)^3(x+y)(x^2-xy+y^2)=945$, respectively. Dividing them (allowed, since neither are $0$), we have \[\frac{xy}{x^2-xy+y^2}=\frac67\] or \[x^2-\frac{13}{6}xy+y^2=0.\] Plugging into the quadratic formula and solving for $x$ in terms of $y,$ we have \[x=\frac{\frac{13y}{6}\pm \sqrt{\frac{169y^2}{36}-4y^2}}{2}=\frac{2y}3 , \frac{3y}2 .\] WLOG, let $x=\frac{3y}2 .$ Plugging into our first equation, we have \[\left(\frac{3}{2}y\right)^4\left(\frac52 y\right)=810 \implies y^3 = 4 .\] Plugging this result (and the one for $x$ in terms of $y$) into our desired expression, we have

\begin{align*} 2x^3+(xy)^3+2y^3 &= \frac{27}{4}y^3 + \left(\frac{3}{2} y^2\right)^3 +2y^3 \\ &= \frac{35}{4}y^3 +\frac{27}{8}(y^3)^2 \\ &= 35+54 \\ &= \boxed{089} \end{align*} ~ASAB

Solution 7

Take $w=x+y$ and $z=xy$. Remark that \begin{align*} ~&2x^{3}+(xy)^{3}+2y^{3} \\ =~&2(x^{3}+y^{3})+(xy)^{3} \\ =~&2\left[(x+y)^{3}-3xy(x+y)\right]+(xy)^{3} \\ =~&2(w^{3}-3wz)+z^{3} \\ =~& 2w^{3}-6wz+z^{3}.\end{align*} The given equations imply that \[wz^{4}=810~~~\text{and}~~~(wz)^{3}-3wz^{4}=945.\] Substituting the first equation into the second, we have that $(wz)^{3}=945+3\cdot 810=3375$, thus $wz=\sqrt[3]{3375}=15$. Now \[z^{3}=\frac{wz^{4}}{wz}=\frac{810}{15}=54\] and \[w^{3}=\frac{3375}{54}=\frac{125}{2}.\] Thus \begin{align*}2w^{3}-6wz+z^{3}&=2\left(\frac{125}{2}\right)-6(15)+54 \\ &=125-90+54 \\ &=\boxed{89}.\end{align*}


Solution 8

$x^4y^4(x+y)=810; x^3y^3(x^3+y^3)=945, \frac{x^2-xy+y^2}{xy}=\frac{7}{6}, \frac{x^2+y^2}{xy}=\frac{13}{6}$

Let $x^2+y^2=13k; xy=6k$, then we can see $(x+y)^2-12k=13k, x+y=5\sqrt{k}$, now, we see $x^4y^4\cdot (x+y)=1296k^4\cdot 5\sqrt{k}, k=\frac{1}{\sqrt[3]{4}}$.

The rest is easy, $2(x^3+y^3)+x^3y^3=216k^3+2[(x+y)^3-3xy(x+y)]=216k^3+2\cdot 35k^{\frac{3}{2}}=\boxed{89}$

~bluesoul

See also

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

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