Indeed easy. But not that ugly. this problem was given in JBMO (2000 i think) Anyway the way to go is not unusual: put $x+y=a$ , $xy=b$ and factorize!!

I think I did it differently but another question: What ages is the JBMO for?

Sorry I got the forum wrong!

riddler wrote: I think I did it differently but another question: What ages is the JBMO for? For kids under 15,5 years old! I took part in this competition when I was in 9th form....... I am willing to see your approach.

riddler wrote: Sorry I got the forum wrong! What do you mean??

socrates wrote: riddler wrote: Sorry I got the forum wrong! What do you mean?? Some moderator moved this post to intermediate forum, dont ask me, I dont know why.

riddler wrote:

Very nice!!

riddler wrote: socrates wrote: riddler wrote: Sorry I got the forum wrong! What do you mean?? Some moderator moved this post to intermediate forum, dont ask me, I dont know why. Which section was it firstly in??

GS forum

riddler wrote: GS forum I think intermediate section is the appropriate place for it!! Although some would consider it pre olympiad........

I moved it. Can you really expect someone who's just getting started with problem solving to be able to solve this?

My 2 cents: this problem is definitely not GS and at least Intermediate. Please take note of this...the difficulty level of the forums is often confused.

Seeing that no understandable solution has been posted, I will post one (for the sake of the Contests Section):

^^^ Perhaps a little clearer in the above: Once you have $x+y=a,xy=b$ AM>GM gives..

riddler wrote: Let $x$ and $y$ be positive reals such that \[ x^3 + y^3 + (x + y)^3 + 30xy = 2000. \] Show that $x + y = 10$. Generalization Let $x,y,k,r$ be positive reals numbers such that \[{{x}^{3}}+{{y}^{3}}+r{{(x+y)}^{3}}+3kxy=\left( r+1 \right){{k}^{3}}\] Show that $x+y=k$.

What I did was I showed 10 would work. Then I showed that $10+a$ and $10-a$ would not work for $a>0$

riddler wrote: Let $x$ and $y$ be positive reals such that \[ x^3 + y^3 + (x + y)^3 + 30xy = 2000. \] Show that $x + y = 10$. We have $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ and $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2- ab - bc - ca)$. From the example we have that $[x^3 + y^3 - 10^3 + 3*10xy] + [(x + y)^3 - 10^3] = 0$,and we apply the formulas and we get $(x + y - 10) * (x^2 + y^2 + (x+y)^2 + 10(x + y) + 200 - 10x - 10y + xy) = 0$ and because the second factor > 0 we obtain $x + y - 10 = 0$ or $x + y = 10$. ionbursuc wrote: riddler wrote: Let $x$ and $y$ be positive reals such that \[ x^3 + y^3 + (x + y)^3 + 30xy = 2000. \] Show that $x + y = 10$. Generalization Let $x,y,k,r$ be positive reals numbers such that \[{{x}^{3}}+{{y}^{3}}+r{{(x+y)}^{3}}+3kxy=\left( r+1 \right){{k}^{3}}\] Show that $x+y=k$. Similar we do the generalisation: $[x^3 + y^3 - k^3 + 3kxy] + r[(x+y)^3 - k^3] = 0$ or $(x + y - k)(x^2+y^2 + 2k^2 - kx - ky +xy + r(x + y)^2 + kr(x+y)) = 0$ and we get $x + y - k = 0$ or $x + y = k$.

We could obviously prove that x+y<10 and Quote: x+y<10 can't be happened.

$2(x^3+y^3-1000)+3xy(x+y-10)=(x+y-10)t$. t is a big expression so I didn't wrote it. But obviously it can't be 0. So x+y=10

Here is a highly non-standard solution to this really easy problem. (The factorization solution is cute but a bit routine) We shall apply the Method of Lagrange Multipliers!!! (Scared eh...? Trust me it's gonna be magical) Indeed, consider the constraint function as \begin{align*} g(x,y)=x^3+y^3+(x+y)^3+30xy-2000 \end{align*} and the objective function \begin{align*} f(x,y)=(x+y-10)^2 \end{align*} We choose to maximize $f$ subject to $g$ and then see that indeed $f \le 0$ but in reality $f \ge 0$ and thus, $x+y=10$ Note that $f,g$ are continuous with continuous partial derivatives. \begin{align*} \nabla f=(2(x+y-10),2(x+y-10)) \end{align*}and also, \begin{align*} \nabla g =(6x^2+3y^2+30y+6xy,6y^2+3x^2+6xy+30x) \end{align*} Evidently, $\nabla g \not= 0$ We define these functions over $U=(0,9000)^2$ for certainly $x,y \in U$. We shall prove that the result holds for any $x,y \ge 0$ satisfying $g$ instead of just positive reals. (This for the informed viewer being just an excuse to execute our notions of topology and compact sets over $\mathbb{R}^2$) Define the closure of the set $U$ as $\bar{U}=[0,9000]^2$. Then consider set $S=\{x \in \bar{U} : g(x)=0\}$ Then, clearly, we have that $S$ is compact and so $f$ achieves a global maxima $x_0$ over $S$. Now, every global maxima is a local one as well and so by the theorem of Lagrange Multipliers Case 1. $x_0 \not \in \bar{U}\cap U$ In this case one of $x,y$ is zero which means that the other is automatically equal to $10$ by our original equation. Case 2. $x_0 \in \bar{U}\cap U$ Thus, we have $\nabla f = \lambda \nabla g$ and therefore we conclude that $2(x+y-10)=\lambda (6x^2+3y^2+6xy+30y)=\lambda (6y^2+3x^2+6xy+30x)$ and so we get $x=y$ or $x+y=10$. While still being ignorant we proceed assuming $x=y$ which from $g$ implies $x=y=5$ and we are done again. Thus, Lagrange Multipliers assure that $f \le 0$ and common sense assures that $f \ge 0$. Thus, $f=0$. $\square$ In my own humble opinion this solution deserves a special prize with a lot of criticism for being overkill. The only insight I intended to pass by writing all this non-sensical stuff was to apply LM in situations where it is unexpected. I would post a note on this soon when I am free on my blog. Do check it out! (Its private as of yet) P.S.- Yay! 400 th post...

I think I hab a shorter solution by case work (seems ironic lol)

Also nice idea there using the $a^3+b^3+c^3-3abc$ factorization (I think it might be getting a bit too slightly overused tho). I will go try it.

This thread started in 2005, then was revived in 2011, then in 2014, 2015, and now 2018.

Also 1st Jan 2016.

No... I see December 31, 2015. Maybe we are in different time zones.

haha! indeed. I come under IST.

Easy problem. Let $x=a+b$ and $y=a-b$. Rewriting the equation and dividing by $2$ gives: $5a^3+3ab^2+15a^2-15b^2=1000$. Therefore $(3a-15)b^2=(1000-15a^2-5a^3)$. However, $3a-15$ is negative in the range $[0,5)$, while $(1000-15a^2-5a^3)$ is positive. Similarly, $3a-15$ is positive for $a>5$, while $1000-15a^2-5a^3$ is negative. This implies that $b^2$ is negative, contradiction. Hence $a=5$, which means $x+y=(a+b)+(a-b)=10$.

Strange solution: Denote $x+y=S$, and $xy=P$. So we get: $3S=\frac{2S^{3}-2000}{S-1}$=$2(S+S+1)-\frac{1998}{S-1}\in Z$, so we must have $S-1\in D_{1998}$ where $1998=2*3^{3}*37$ and testing these, we found $S=10$

Note that $x^3+y^3+(x+y)^3+30xy-2000=0$ factors as $(x+y-10)(2x^2+xy+2y^2+20x+20y+200)=0$ If $x,y$ are both positive then the second factor is positive, so that we are forced to conclude $x+y=10$.

After some manipulation we get: $3xy(10 - (x+y)) = 2((x+y)^3 - 10^3)$ Case 1: $x+y > 10:$ $RHS > 0$, So $LHS$ has to be $> 0$, so $10 - (x+y) > 0$ or $x+y < 10 => contradiction!$ Case 2: $x+y < 10:$ $RHS < 0$, So $LHS$ has to be $< 0$, so $10 - (x+y) < 0$ or $x+y > 10 => contradiction!$ Thus $x + y = 10$

AM GM Solution!!!!!!! We can factor our given expression into $2(x+y)^3-3xy(x+y)+30xy=2000$. Let $x+y=a, xy=b$. Then we get that $2a^3-3ab+30b=2000$. Solving for b, we get that $b=\frac{2a^3-2000}{3a-30}$. If a is not equal to 10, this can be simplified as $b=\frac{2a^2+20a+200}{3}$. By AM-GM, we know that $\frac{a^2}{4} \ge b$. When we plug in what we found for b, we get that this simply cannot occur, as the inequality won't hold when a is positive. Now let us see if a=10 works. Then we get b=0, meaning that our possible ordered pairs of $(x, y)$ are $(10, 0), (0, 10)$, which indeed do work.

Here's my work. Rearrangement gives $$x^3+y^3+(x+y)^3+30xy-2000=0.$$Then, let $a=x+y, b=xy.$ We get $$x^3+y^3+a^3+30b-2000=0 \implies (x+y)^3-3x^2y-3xy^2+a^3+30b-2000=0 \implies a^3-3(by+bx)+a^3+30b-2000=0 \implies a^3-3ab+30b+a^3-2000=0.$$Applying Difference of Cubes, $$2(a-10)(a^2+10a+100)+3a+30b=0.$$Then, substituting back $a=x+y, b=xy,$ we have $$2(x+y-10)[(x+y)^2+10(x+y)+100]-3xy(x+y) \implies (x+y-10)(2x^2+2y^2+xy+20x+20y+200)=0. ~ ~ ~ \text{(By factoring)}$$Clearly, the bracket on the right is non-zero, and the desired result follows. (Because $x+y-10=0$ implies that $x+y=10.$) Feel free to give any constructive feedback below! I would really appreciate it!

KRIS17 wrote: After some manipulation we get: $3xy(10 - (x+y)) = 2((x+y)^3 - 10^3)$ Case 1: $x+y > 10:$ $RHS > 0$, So $LHS$ has to be $> 0$, so $10 - (x+y) > 0$ or $x+y < 10 => contradiction!$ Case 2: $x+y < 10:$ $RHS < 0$, So $LHS$ has to be $< 0$, so $10 - (x+y) < 0$ or $x+y > 10 => contradiction!$ Thus $x + y = 10$ You can factor the RHS as a difference of two cubes and then you can divide both sides by 10-xy which then leads to a contradiction so 10-xy has to be 0.

Why have I seen this before somewhere?

Denote $a=x+y,b=xy$. The equation is: $$2a^3-3ab+30b=2000$$$$\Leftrightarrow2(a^3-1000)-3b(a-10)=0$$$$\Leftrightarrow(a-10)(2a^2+20a+200-3b)=0$$Assume FTSOC that $a\ne10$, so $2a^2+20a+200=3b$. Then by AM-GM, $a\ge2\sqrt b$, so $2a^2+20a+200\ge8b+40\sqrt b+200>3b$, contradiction reached. $\square$

Let $a=x+y$ and $b=xy$. We wish to prove $a=10$. We have $a(a^2 - 3b) + a^3 + 30b = 2000$, so $2a^3 - 3ab +30b = 2000$. Now, $2a^3 - 3ab + 30b - 2000 = (a-10)(2a^2 + 20a + 200 - 3b)=0$. Suppose FTSOC $a\ne 10$. Then $2a^2 + 20a + 200 = 3b$. Note that since $(x+y)^2 = x^2+y^2+2xy \ge 4xy$, $a^2 \ge 4b$. So $2a^2 + 2a + 200 \ge 4b + 2a + 200 > 3b$, contradiction. So $a=10$.