mhet49 wrote: Find all functions $f:R\to R$ such that f($x^3$) + f($y^3$) = (x+y)f($x^2$)+ f($y^2$) - f(xy) for all x $\in$ R. Take $x=y=0$ than $f(0)=0$ If $x=0$ than $f(y^3)=f(y^2)$ ..(a) If $y=0$ than $f(x^3)=xf(x^2)$ or $f(y^3)=yf(y^2)$ than for all $z=y^2\geq 0$ we have $f(z)=0$ and from ..(a) $f(x)=0$ ,for all $x\in R$

Bravo Arber.It is easy .I got the same result but i used a bit a different method

mhet49, can you please confirm you meant what you posted? You're actual post was:

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Which I edited to:

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I have a feeling the problem should have been: mhet49 wrote: Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that \[f(x^3)+f(y^3)=(x+y)(f(x^2)+f(y^2)- f(xy))\] for all $x\in\mathbb{R}$.

Anyway, here is my solution to problem that I think is intended. Let $P(x,-x)$ be the assertion that $f(x^{3})+f(y^{3})=(x+y)(f(x^{2})+f(y^{2})-f(xy))$. Then $P(\sqrt[3]{x},\sqrt[3]{-x})\implies f(x)=-f(-x)$. Also, $P(0,0)\implies f(0)=0$. Therefore, $P(x,0)\implies f(x^3)=xf(x^2)$. Expanding $f(x^{3})+f(y^{3})=(x+y)(f(x^{2})+f(y^{2})-f(xy))$ and using $f(x^3)=xf(x^2)$, we see that $(x+y)f(xy)=xf(y^2)+yf(x^2)$ which we will call the assertion $Q(x,y)$. Then, $Q(x,1)\implies (x+1)f(x)=xf(1)+f(x^2)$ and $Q(x,-1)\implies (x-1)(-f(x))=xf(1)-f(x^2)$ (since $f(x)=-f(-x)$). Adding these two equations implies that $2f(x)=2xf(1)$ which gives the solution $f(x)=cx$ for any real constant $c$.

WakeUp wrote: I have a feeling the problem should have been: mhet49 wrote: Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that \[f(x^3)+f(y^3)=(x+y)(f(x^2)+f(y^2)- f(xy))\] for all $x\in\mathbb{R}$. In this case, we can solve it as follows: Let $x=y=0$ to see $f(0)=0$. Then $P(x, 0)\implies f(x^3)=xf(x^2)$ and $P(x, -x)\implies f(n)=-f(-n)$ for all real $n$. Then $P(x, -x)\implies f(x^3)-f(y^3)=(x-y)(f(x^2)+f(y^2)+f(xy))$. Adding this to the original and using $f(x^3)=xf(x^2)$ gives $yf(xy)=xf(y^2)$, and plugging $y=1$ into this gives $f(x)=xf(1)=ax$ for a constant $a$. We see that this always works, so $\boxed{f(x)=xf(1)}$ is the solution. Edit: beaten, but this solution is a little different. Edit2: Wow, this matches up exactly with the solution in the below link..

As I was reading your solution, I realised I found it familiar.. That's when I realised this problem is from the British Maths Olympiad Round 2 2009 Posted at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=254396, for example.

i have use the same way as u.But in the end we can find that f(1)=0.

WakeUp wrote: mhet49, can you please confirm you meant what you posted? You're actual post was:

Click to reveal hidden text

Which I edited to:

Click to reveal hidden text

I have a feeling the problem should have been: mhet49 wrote: Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that \[f(x^3)+f(y^3)=(x+y)(f(x^2)+f(y^2)- f(xy))\] for all $x\in\mathbb{R}$. Actually no, it is just like what I have written it.

mhet49 wrote: Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that \[f(x^3)+f(y^3)=(x+y)f(x^2)+f(y^2)- f(xy)\] for all $x\in\mathbb{R}$. My solution was as follows: f(x^{3})+ f(y^{3})= (x+y)f(x^{2})+ f(y^{2})- f(xy) the eqn for x=0 and y=o gives us f(0)=0 which we'll use for getting many valuable equations: first plug x=0, then we'll have f(y^{3})= f(y^{2}) whilst by plugging y=0 we get f(x^{3})= x f(x^{2}) so it means that f(x^{2})=0; also in the eqn for y=1 we have f(x)=f(x^{2}) which means f(x)=0 is the solution.

mhet49 wrote: Actually no, it is just like what I have written it. Then I bet the proposers messed up the bracketing - can you ask them?

mavropnevma wrote: mhet49 wrote: Actually no, it is just like what I have written it. Then I bet the proposers messed up the bracketing - can you ask them? They had!! This is what we had in the competition and what they corrected was with bracket like u said.... Unfortunate us

Setting $x=y=$ in the equation we get $f(0)=0$. Then setting $y=0$ we get $f(x^3)=xf(x^2)$; in particular, putting $x \mapsto -x$ and adding we get $f(-x^3)=-f(x^3)$ so $f$ is odd. Finally, substituting $y \mapsto -y$ in the original equation and then adding it to the original equation we have $yf(xy)=xf(y^2)$, which for $y=1$ gives $f(x)=f(1) x$. Conversely, it is easily verified that, for any $c \in \mathbb R$, $f(x)=cx$ satisfies the conditions.

Cassius, that is the exact same solution as applepi2000's, there's really no need to post it again.. you've also done this in a couple of other threads

Dear WakeUp, I first read the problems on the resources page and solve them, then post my solutions on the threads and read others' after, so it may happen that a similar one was already posted. Moreover, I do it primarily also as an exercise to learn to write solutions since I have much need for it, so I will be glad if you will point out any mistake or omissions in mine. Having said that, I will avoid to post solutions to problems already solved, and please warn me if I do that error again.

Well it seems fair enough, I don't mean to nag or be annoying, feel free to do so if you wish. But, for example, in this case, each step of your proof is identical to applepi2000's, and there is not much point spamming a thread with the same proof (it reminds me of the very first IMO problem, where nearly 10 equivalent proofs have been posted... thankfully that thread has been locked now though. In fact, it might have been me that locked it, I can't remember ). I suppose it is good practice for writing up solutions... but if there is already an identical proof posted, maybe try to avoid it! An alternative could be to post it on your blog or something..

So $f(x^3)+f(y^3)=(x+y)(f(x^2)+f(y^2)-f(xy))$. Plugging in $x=0$, $f(0)+f(y^3)=yf(0)+yf(y^2)-yf(0)=yf(y^2)$. Now, $y=0$ yields $2f(0)=0$, so $f(0)=0$. Thus, $f(y^3)=yf(y^2)$, and $f(-y^3)=-yf(y^2)$. Therefore, $f(x)$ is odd (we could have found this with $y=-x$, but whatever). Thus, $xf(x^2)+yf(y^2)=xf(x^2)+yf(y^2)+yf(x^2)+xf(y^2)-xf(xy)-yf(xy)$. Then, $yf(x^2)+xf(y^2)=(x+y)f(xy)$. Letting $y=1$, $f(x^2)+xf(1)=(x+1)f(x)$. Also, $y=-1$ yields $-f(x^2)+xf(1)=(x-1)f(-x)$. Adding the two, $f(x^2)+f(-x^2)+xf(1)+xf(1)=2xf(1)=xf(x)+xf(-x)+f(x)-f(-x)=2f(x)$. Then $2f(x)=2xf(1)$, so $f(x)=xf(1)$. Thus, $f(x)=ax$. We have $yf(x^2)+xf(y^2)=(x+y)f(xy)$, so $yf(x^2)+xf(y^2)=ax^2y+axy^2=axy(x+y)=(x+y)f(xy)$. Thus, the solutions are all of the form $f(x)=ax$, and anything of that form is a solution. Q.E.D.

My solution (WITHOUT BRACKETS)..... Putting $x$ = $y$ = $0$, we get $f(0) = 0$. Putting $y$ = $0$, we get $f(x^3) = xf(x^2)$. Putting $x$ =$0$, we get $f(y^3) = f(y^2)$. Solving, we get $f(y^2) = 0$, provided $y$ ≠ $1$. Putting $y$ = $1$, we get $f(x^2) = f(x)$. This implies that $f$ is even. So, $f(1) = f(-1) = 0$. Hence, $f(x) = 0$ for all $x\in\mathbb{R}$.

Plugging in $x=y=0$, we have that $2f(0)=f(0)-f(0) \implies f(0)=0$. Plugging in $x=0$, we have $f(0)+f(y^3)=(y)f(0)+f(y^2)- f(0) \implies f(y^3)=f(y^2)$. Plugging in $y=0$, we see that $f(x^3)+f(0)=(x)f(x^2)+f(0)- f(0) \implies f(x^3)= x{f(x^2)}$. Comparing the last two statements, we see have that $x{f(x^2)}=f(x^3)=f(x^2) \implies x{f(x^2)}=f(x^2)$, implying that x is always zero, or that $f(x)$ is always zero, of which the former is obviously false, hence $f(x)=0$. $\square$

Let $P(x,y)$ be the assertion that $f(x^{3})+f(y^{3})=(x+y)(f(x^{2})+f(y^{2})-f(xy))$. $P(x,-x),f(x)+f(-x)=0$ so $f(0)=0$ $P(x,0),f(x^3)=xf(x^2)$ and $P(x,y) \implies (x+y)f(xy)=yf(x^2)+xf(y^2)$ $(x+y)f(xy)=yf(x^2)+xf(y^2)$ above set $y=1,(x+1)f(x)=f(x^2)+xf(1)$ and set $y=-1,(x-1)f(x)=-f(x^2)+xf(1) \implies f(x)=xf(1)$ for all $x \in R$

MonsterS wrote: Let $P(x,y)$ be the assertion that $f(x^{3})+f(y^{3})=(x+y)(f(x^{2})+f(y^{2})-f(xy))$. $P(x,-x),f(x)+f(-x)=0$ so $f(0)=0$ $P(x,0),f(x^3)=xf(x^2)$ and $P(x,y) \implies (x+y)f(xy)=yf(x^2)+xf(y^2)$ (x+y)f(xy)=yf(x^2)+xf(y^2) above set $y=1,(x+1)f(x)=f(x^2)+xf(1)$ and set $y=-1,(x-1)f(x)=-f(x^2)+xf(1) \implies f(x)=xf(1)$ for all $x \in R$ very nice solution

$x=0=y\implies f(0)=0$ $x=y\implies f(x^3)=xf(x^2)$ So coming back to the first equation $xf(x^2)+yf(y^2)=(x+y)f(x^2)+f(y^2)-f(xy)$ Put above $y=1$ and get $f(x^2)=f(x)$ Now put to the given equation $y=0$ and get $f(x^3)=f(x^2)$ So for all $x\in R$ we have $f(x^2)=xf(x^2)\iff 0=(x-1)f(x^2)\iff 0=(x-1)f(x) $ That means $x\in R\wedge x\neq 1\implies f(x)=0$ Using $f(x^3)=xf(x^2)$ put $x=-1$ and get $0=f(-1)=-f(1)\implies f(1)=0$ So the solution is $\forall x\in R\ f(x)=0$. It's easy to check that it satisfies given conditions.

Ez. Note that $x=y=0$ gives $f(0)=0$. Take $y=0$ to get $f(x^3)=xf(x^2)$. Additionally, $x=0$ gives $f(y^3)=f(y^2)$. Recalling $f(y^3)=yf(y^2)$, we get $f(y^2)=0$ for all $y\neq 1$. Thus, $f(y^3)=yf(y^2)=0$, which means $f=0$ for all $x\neq 1$. Now take $P(1,2)$ to get $f(1)=0$ as well. Thus $\boxed{f=0}$

By applying symmetry we see that x+y=1 and then we get the only solution as f(x) =0

$P(0,0): f(0)=0$ $P(x,0): f(x^3)=xf(x^2)$ $P(0,y): f(y^3)=f(y^2)$ so $f(x^3)=xf(x^3)$ and since any real number can be represented as a cube of another real, we must have $f(x)=0$ for all real $x$ except $1$, but since $f$ is even $f(1)=0$ follows from $0=f(-1)=f(1)$ and so $f(x)=x$ for all real $x$.

$P(0,0)\Rightarrow f(0)=0$ $P(x,0)\Rightarrow f(x^3)=xf(x^2)$ $P(0,x)\Rightarrow f(x^3)=f(x^2)\Rightarrow (1-x)f(x^2)=0$, so $f(x)=0\forall x\in[0,1)\cup(1,\infty)$. Also, note that $f$ is even, so it suffices to check $f(1)=0$. This holds because of $P(1,-1)$.

Let $P(x,y)$ denote the assertion $f(x^3)+f(y^3)=(x+y)(f(x^2)+f(y^2)-f(xy)).$ $P(x,x)\implies f(x^3)=xf(x^2)$ $P(x,-x)\implies -f(x)=f(-x)$ Combining $P(x,1)$ and $P(x,-1)$ $\implies f(x)=cx$ which works.

Can someone please confirm what the answer is?? Half of solutions get f(x)=0 and half of them f(x)=kx and I am very very confused

postingunsolvedquestions wrote: Can someone please confirm what the answer is?? Half of solutions get f(x)=0 and half of them f(x)=kx and I am very very confused Maybe because half of posts solve original problem $f(x^3)+f(y^3)=(x+y)f(x^2)+f(y^2)- f(xy)$ While other half solved modified problem $f(x^3)+f(y^3)=(x+y)(f(x^2)+f(y^2)- f(xy))$

mhet49 wrote: Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that \[f(x^3)+f(y^3)=(x+y)f(x^2)+f(y^2)- f(xy)\]for all $x\in\mathbb{R}$. $x=y=0$ we have $f(0)=0$ $x=0$ we have $f(x^3)=f(x^2)$ $y=0$ we have $f(x^3)=xf(x^2)$ $\Rightarrow f(x)=0, \forall x \ge 0$ Replace: $-f(xy)=0 \Rightarrow f(x)=0$, $ \forall x \in \mathbb{R}$

f(x^3)+f(y^3) = (x+y)f(x^2)+f(y^2)-f(xy) P(x,x) -> 2f(x^3) = 2xf(x^2) -> f(x^3) = xf(x^2) ... (i) P(y,y) -> 2f(y^3) = 2yf(y^2) -> f(y^3) = yf(y^2) ... (ii) P(0,y) -> f(y^3) = f(y^2) -> P(x) = f(x^3) = f(x^2) ... (iii) Comparing (i), (ii) and (iii), f(x^2) = f(x^3) = xf(x^2) -> f(x^2)-xf(x^2) = 0 -> f(x^2)(1-x) = 0 Either x = 1 or f(x) = 0 Since, there x is an independent constant, f(x) is 0.

Might be the same as many solution before $P(0,0)\Longrightarrow f(0)=0$ $P(x,0)\Longrightarrow f(x^3)=xf(x^2)$ $P(0,x)\Longrightarrow xf(x^2)=f(x^3)=f(x^2)$ so $f(x^2)=0\Longrightarrow f(x^3)=0\Longrightarrow f(x)=0$

Answer : f(x) = 0. Solution : Let $ P(x,y) $ denote the given assertion. $ P(0,0) \Rightarrow f(0) = 0 $. $ P(x,0) \Rightarrow f(x^3) = xf(x^2). P(x,1) \Rightarrow f(x^2) = f(x) $. $ x \leftrightarrow y , y \to 0 \Rightarrow xf(x) = f(x) \Rightarrow f(x) = 0 $.

mhet49 wrote: Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that \[f(x^3)+f(y^3)=(x+y)f(x^2)+f(y^2)- f(xy)\]for all $x\in\mathbb{R}$. OK, so assuming this is the problem my solution is as follows. First, $P(0,0)$ gives us that $f(0)=0$. Then, \[P(x,0) \implies f(x^3)=xf(x^2)\]\[P(0,x) \implies f(x^3)=f(x^2)\]for all $x\in \mathbb{R}$. Combining these two relations we have that \[xf(x^2)=f(x^3)=f(x^2)\]for all $x\in \mathbb{R}$. Thus, $f(x^2)=0$ for all $x\neq 1$. But this means that $f(x^2)=0$ for all $x$ (we can consider $x$ to be -1 to deal with the edge case when $x=1$). Thus, $f(t)=0$ for all $t\geq 0$. Then, for all $x\geq 0$, \[P(-x,-x)\implies 2f(-x^3)=-2f(x^2)+f(x^2)-f(x^2)=-2f(x^2)\]which implies that $f(-x^3)=-f(x^2)$ for all $x\geq 0$. Thus, $f(t)=0$ for all $t \leq 0$ as well. In conclusion the only working solution is $f\equiv 0$ and we are done.

Let $P(x,y): f(x^3)+f(y^3) = (x+y)f(x^2)+f(y^2)-f(xy)$. Now, we take $P(0,0): 2f(0) = f(0)-f(0) = 0$, we obtain $f(0)= 0$ $P(0,y): f(y^3) = f(y^2)$ In this case we can obtain taking $y = -1$ $f(-1) = f(1)\ldots(1)$ $P(x,0): f(x^3) = xf(x^2)$, also taking $y=-1$, we obtain $f(-1) = -f(1)$, from $(1)$,$-f(1) = f(1)\Rightarrow f(1) = 0$ Now, \[f(x^2) = f(x^3) = xf(x^2)\]So $0=(x-1)f(x^2)\forall x \in \mathbb{R}$, if I take $x\neq 1$, so $f(\alpha) = 0 \forall 1\neq \alpha\geq 0 $ but f(1) is also 0, so $f(\alpha) = 0$ for all $\alpha \geq 0 $ remembering $f(x^3)=f(x^2) = 0$ because $x^2 \geq 0 $ for all $x$ real number. $x^3$ goes through all the reals Therefore $f\equiv 0$. As desired.

Corrected version wrote: Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that \[f(x^3)+f(y^3)=(x+y)(f(x^2)+f(y^2)- f(xy))\]for all $x\in\mathbb{R}$. Sketch: $P(x, x)$ for $f(x^3) = xf(x^2)$, $P(x, -y) + P(x, y)$ works.

$P(x,x)$ $gives$ $===>$ $f(x^3)=x*f(x^2)$ $(1)$ Then from $P(0,0)$ $===>$ $f(0)=0$ $(2)$ $P(0,x)$ $===>$ $f(x^3)=f(x^2)$ $(3)$ $(3)$ $and$ $(1)$ $finishes$ $the$ $problem$ $f(x^3)=f(x^2)=x*f(x^2)$ ofc x is not a constant , so $f(x)=0$!