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

Latest revision as of 21:21, 26 November 2023

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 1

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 $2(x^3+y^3)=1890+x^6y^6,$ 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

Solution 9

Let's first put the left hand sides of the equations in factored forms. Doing this we obtain $(xy)^4 \cdot (x + y) = 810$ and $(xy)(x + y)(x^2 - xy + y^2) = 945$ . Now, we will subtract and add the equations to gather information on $x$ and $y$ . When we subtract the equations and clean it up via factoring, we yield $(xy)^3 \cdot (x + y) \cdot (x - y)^2 = 115$ , and when we add them, we yield $(xy)^3 \cdot (x + y) \cdot (x^2 + y^2) = 1755$ . Now with some intuition, you should divide the equations to obtain $\frac{(x^2 + y^2)}{(x - y)^2} = 13$ . Now, we clean this up to obtain the following factoring of $0 = 2 \cdot (2x - 3y) \cdot (3x - 2y)$ . This implies that $x = \frac{3y}{2}$ . We plug that into the target expression to reduce it down to one variable, and get that target expression is $2x^3 + (xy)^3 + 2y^3 = \frac{27}{4} \cdot y^3 + \frac{27}{8} \cdot y^6 + 2y^3$ . This means that if we can find a way to get $y^3$ , then the rest is trivial. We get $y^3$ by plugging in $x = \frac{3y}{2}$ into $x^3 \cdot y^6 + y^3 \cdot x^6 = 945$ . However, this time we only factor as $(xy)^3 \cdot (x^3 + y^3)$ because we particularly want a cubic degree on $y$ . Plugging in $x = \frac{3y}{2}$ we get $y^3 = 4$ . Now lets plug this into our target expression to get $\frac{27}{4} \cdot 4 + \frac{27}{8} \cdot 4^2 + 2 \cdot 4 = \boxed{89}$ .

~triggod, $\LaTeX$ by MinecraftPlayer404

Solution 10 (Newton sums)

Let $x, y$ be roots of some polynomial $P(\alpha)$ . Then we have that $P(\alpha) = (\alpha - x)(\alpha - y) \implies P(\alpha) = \alpha^2 - (x + y) \alpha + xy$ . Now from the first equation we have that $x + y = 810/(xy)^4$ . For convenience, denote $z = xy$ .

Define $P_k = x^k + y^k$ , where $x, y$ are roots of the above polynomial. Then we have that by Newton's formulas $P_k = (x + y)P_{k - 1} - xyP_{k - 2} = 810/z^4P_{k - 1} - zP_{k - 2}$ , where $P_0 = 2$ and $P_1 = x + y = 810/z^4$ . We desire $2P_3 + z^3$ . Now building up this recurrence;

$P_2 = \frac{810}{z^4}P_1 - zP_0 = \frac{810^2}{z^8} - 2z.$ Then we have that $P_3 = \frac{810}{z^4} \left(\frac{810^2}{z^8} - 3z \right).$ Plugging this value of $P_3$ into the second equation $z^3P_3 = 810$ , yields $z^9 = 164025$ . Now $z^3 = 54$ . To compute $P_3$ is now trivial, as returning this new value of $z$ into the second equation yields $P_3 = 35/2$ . Hence $z^3 + 2P_3 = 54 + 35 = \boxed{089}$ .

-th1nq3r

Video Solution by mop 2024

https://youtu.be/7VHEGNRQKeI

~r00tsOfUnity

@@ Line 3: / Line 3: @@
 Let <math>x</math> and <math>y</math> be real numbers satisfying <math>x^4y^5+y^4x^5=810</math> and <math>x^3y^6+y^3x^6=945</math>. Evaluate <math>2x^3+(xy)^3+2y^3</math>.
-==Solution==
+==Solution 1==
 The expression we want to find is <math>2(x^3+y^3) + x^3y^3</math>.
 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{089}</math>.
+Note that since the value we want to find is <math>2(x^3+y^3)+x^3y^3</math>, we can convert <math>2(x^3+y^3)</math> into an expression in terms of <math>x^3y^3</math>, since from the second equation which is <math>x^3y^3(x^3+y^3)=945</math>, we see that <math>2(x^3+y^3)=1890+x^6y^6,</math> and thus the value is <math>\frac{1890+x^6y^6}{x^3y^3}.</math> Since we've already found <math>x^3y^3,</math> we substitute and find the answer to be 89.
 ==Solution 2==
@@ Line 17: / Line 19: @@
 ==Solution 4==
-As with the other solutions, factor. But this time, let <math>a=xy</math> and <math>b=x+y</math>. Then <math>a^4b=810</math>. Notice that <math>x^3+y^3 = (x+y)(x^2-xy+y^2) = b(b^2-3a)</math>. Now, if we divide the second equation by the first one, we get <math>7/6 = \frac{b^2-3a}{a}</math>; then <math>\frac{b^2}{a}=\frac{25}{6}</math>. Therefore, <math>a = \frac{6}{25}b^2</math>. Substituting <math>a</math> into <math>b</math> in equation 2 gives us <math>b^3 = \frac{5^3}{2}</math>; we are looking for <math>2b(b^2-3a)+a^3</math>. Finding <math>a</math>, we get <math>35</math>. Substituting into the first equation, we get <math>b=54</math> <math>Our final answer is 35+54=\boxed{089}</math>.
+As with the other solutions, factor. But this time, let <math>a=xy</math> and <math>b=x+y</math>. Then <math>a^4b=810</math>. Notice that <math>x^3+y^3 = (x+y)(x^2-xy+y^2) = b(b^2-3a)</math>, so we are looking for <math>2b(b^2-3a)+a^3</math>. Now, if we divide the second equation by the first one, we get <math>7/6 = \frac{b^2-3a}{a}</math>; then <math>\frac{b^2}{a}=\frac{25}{6}</math>. Therefore, <math>a = \frac{6}{25}b^2</math>. Substituting <math>\frac{6}{25}b^2</math> for <math>a</math> in equation 1, simplifying, and then taking the cube root gives us <math>b^3 = \frac{5^3}{2}.</math> Finding <math>a^3</math> by cubing <math>a = \frac{6}{25}b^2</math> on both sides and simplifying using our previous substitution, we get <math>a^3 = 54</math>. Substituting this into the first equation and then dividing by <math>27</math>, we get <math>2b(b^2 - 3a) = 35</math>. Our final answer is <math>35+54=\boxed{089}</math>.
+==Solution 5==
+Factor the given equations as:
+<cmath>x^4y^4(x+y)=810</cmath>
+<cmath>x^3y^3(x^3+y^3)=x^3y^3(x+y)(x^2-xy+y^2)=945</cmath>
+We note that these expressions (as well as the desired expression) can be written exclusively in terms of <math>x+y</math> and <math>xy</math>. We make the substitution <math>s=x+y</math> and <math>p=xy</math> (for sum and product, respectively).
+<cmath>x^4y^4(x+y)=p^4s=810</cmath>
+<cmath>x^3y^3(x+y)(x^2-xy+y^2)=p^3(s)(s^2-3p)=s^3p^3-3p^4s=945</cmath>
+We see that <math>p^4s</math> shows up in both equations, so we can eliminate it and find <math>sp</math>, after which we can get <math>p^3</math> from the first equation. If you rewrite the desired expression using <math>s</math> and <math>p</math>, it becomes clear that you don't need to actually find the values of <math>s</math> and <math>p</math>, but I will do so for the sake of completion.
+<cmath>s^3p^3=945+3p^4s</cmath>
+<cmath>s^3p^3=945+3(810)=3375</cmath>
+<cmath>sp=15</cmath>
+<cmath>p^3=\frac{810}{sp}=54</cmath>
+<cmath>p=3\cdot2^{1/3}</cmath>
+<cmath>s=\frac{15}{p}=5\cdot2^{-1/3}</cmath>
+The desired expression can be written as:
+<cmath>2(x^3+y^3)+(xy)^3=2(x+y)(x^2-xy+y^2)+(xy)^3</cmath>
+<cmath>2(s)(s^2-3p)+p^3=2s^3-6sp+p^3</cmath>
+Plugging in <math>s</math> and <math>p</math>, we get:
+<cmath>2(5\cdot2^{-1/3})^3-6(15)+54=125-90+54=\boxed{089}</cmath>
+- gting
+==Solution 6==
+Factor the first and second equations as <math>(xy)^4(x+y)=810</math> and <math>(xy)^3(x+y)(x^2-xy+y^2)=945</math>, respectively. Dividing them (allowed, since neither are <math>0</math>), we have
+<cmath>\frac{xy}{x^2-xy+y^2}=\frac67</cmath>
+or
+<cmath>x^2-\frac{13}{6}xy+y^2=0.</cmath>
+Plugging into the quadratic formula and solving for <math>x</math> in terms of <math>y,</math> we have
+<cmath>x=\frac{\frac{13y}{6}\pm \sqrt{\frac{169y^2}{36}-4y^2}}{2}=\frac{2y}3 , \frac{3y}2 .</cmath>
+WLOG, let <math>x=\frac{3y}2 .</math>
+Plugging into our first equation, we have
+<cmath>\left(\frac{3}{2}y\right)^4\left(\frac52 y\right)=810 \implies y^3 = 4 .</cmath>
+Plugging this result (and the one for <math>x</math> in terms of <math>y</math>) into our desired expression, we have
+<cmath>
+\begin{align*}
+x^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*}
+</cmath>
+~ASAB
+==Solution 7==
+Take <math>w=x+y</math> and <math>z=xy</math>. Remark that <cmath>\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*}</cmath> The given equations imply that <cmath>wz^{4}=810~~~\text{and}~~~(wz)^{3}-3wz^{4}=945.</cmath> Substituting the first equation into the second, we have that <math>(wz)^{3}=945+3\cdot 810=3375</math>, thus <math>wz=\sqrt[3]{3375}=15</math>. Now <cmath>z^{3}=\frac{wz^{4}}{wz}=\frac{810}{15}=54</cmath> and <cmath>w^{3}=\frac{3375}{54}=\frac{125}{2}.</cmath> Thus <cmath>\begin{align*}2w^{3}-6wz+z^{3}&=2\left(\frac{125}{2}\right)-6(15)+54 \\ &=125-90+54 \\ &=\boxed{89}.\end{align*}</cmath>
+==Solution 8==
+<math>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}</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>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
+==Solution 9==
+Let's first put the left hand sides of the equations in factored forms. Doing this we obtain <math>(xy)^4 \cdot (x + y) = 810</math> and <math>(xy)(x + y)(x^2 - xy + y^2) = 945</math>. Now, we will subtract and add the equations to gather information on <math>x</math> and <math>y</math>. When we subtract the equations and clean it up via factoring, we yield <math>(xy)^3 \cdot (x + y) \cdot (x - y)^2 = 115</math>, and when we add them, we yield <math>(xy)^3 \cdot (x + y) \cdot (x^2 + y^2) = 1755</math>. Now with some intuition, you should divide the equations to obtain <math>\frac{(x^2 + y^2)}{(x - y)^2} = 13</math>. Now, we clean this up to obtain the following factoring of <math>0 = 2 \cdot (2x - 3y) \cdot (3x - 2y)</math>. This implies that <math>x = \frac{3y}{2}</math>. We plug that into the target expression to reduce it down to one variable, and get that target expression is <math>2x^3 + (xy)^3 + 2y^3 = \frac{27}{4} \cdot y^3 + \frac{27}{8} \cdot y^6 + 2y^3</math>. This means that if we can find a way to get <math>y^3</math>, then the rest is trivial. We get <math>y^3</math> by plugging in <math>x = \frac{3y}{2}</math> into <math>x^3 \cdot y^6 + y^3 \cdot x^6 = 945</math>. However, this time we only factor as <math>(xy)^3 \cdot (x^3 + y^3)</math> because we particularly want a cubic degree on <math>y</math>. Plugging in <math>x = \frac{3y}{2}</math> we get <math>y^3 = 4</math>. Now lets plug this into our target expression to get <math>\frac{27}{4} \cdot 4 + \frac{27}{8} \cdot 4^2 + 2 \cdot 4 = \boxed{89}</math>.
+~triggod, <math>\LaTeX</math> by MinecraftPlayer404
+==Solution 10 (Newton sums)==
+Let <math>x, y</math> be roots of some polynomial <math>P(\alpha)</math>. Then we have that <math>P(\alpha) = (\alpha - x)(\alpha - y) \implies P(\alpha) = \alpha^2 - (x + y) \alpha + xy</math>. Now from the first equation we have that <math>x + y = 810/(xy)^4</math>. For convenience, denote <math>z = xy</math>.
+Define <math>P_k = x^k + y^k</math>, where <math>x, y</math> are roots of the above polynomial. Then we have that by Newton's formulas <math>P_k = (x + y)P_{k - 1} - xyP_{k - 2} = 810/z^4P_{k - 1} - zP_{k - 2}</math>, where <math>P_0 = 2</math> and <math>P_1 = x + y = 810/z^4</math>. We desire <math>2P_3 + z^3</math>. Now building up this recurrence;
--gorefeebuddie
+<cmath>P_2 = \frac{810}{z^4}P_1 - zP_0 = \frac{810^2}{z^8} - 2z.</cmath> Then we have that <cmath>P_3 = \frac{810}{z^4} \left(\frac{810^2}{z^8} - 3z \right). </cmath> Plugging this value of <math>P_3</math> into the second equation <math>z^3P_3 = 810</math>, yields <math>z^9 = 164025</math>. Now <math>z^3 = 54</math>. To compute <math>P_3</math> is now trivial, as returning this new value of <math>z</math> into the second equation yields <math>P_3 = 35/2</math>. Hence <math>z^3 + 2P_3 = 54 + 35 = \boxed{089}</math>.
+-th1nq3r
+==Video Solution by mop 2024==
+https://youtu.be/7VHEGNRQKeI
+~r00tsOfUnity
 ==See also==
 {{AIME box|year=2015|n=II|num-b=13|num-a=15}}
 {{MAA Notice}}
+[[Category:Intermediate Algebra Problems]]

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