Set $ y=0$ then $ f(x^3)=xf(x^2)$ , so $ f(y^3)=yf(y^2)$ , then $ f(x^3+y^3)=xf(x^2)+yf(y^2)=f(x^3)+f(y^3)$ $ \rightarrow$ $ f(x)+f(y)=f(x+y)$ simple induction gives : $ f(x_1+x_2+...+x_n)=f(x_1)+f(x_2)+f(x_3)+...+f(x_n)$ so $ \forall \ n\in\mathbb{R}$ then $ f(nx)=nf(x)$ and thus by setting $ x=1$ we have $ f(n)=nf(1)$, since $ f(1)$ is a constant then by assuming $ f(1)=a$ , gives $ f(n)=an$ with $ a\in\mathbb{R}$ done

matrix41 wrote: Set $ y = 0$ then $ f(x^3) = xf(x^2)$ , so $ f(y^3) = yf(y^2)$ , then $ f(x^3 + y^3) = xf(x^2) + yf(y^2) = f(x^3) + f(y^3)$ $ \rightarrow$ $ f(x) + f(y) = f(x + y)$ simple induction gives : $ f(x_1 + x_2 + ... + x_n) = f(x_1) + f(x_2) + f(x_3) + ... + f(x_n)$ so $ \forall \ n\in\mathbb{R}$ then $ f(nx) = nf(x)$ Why for all $ n \in R$ : matrix41 wrote: Set $ y = 0$ then $ f(x^3) = xf(x^2)$ , so $ f(y^3) = yf(y^2)$ , then $ f(x^3 + y^3) = xf(x^2) + yf(y^2) = f(x^3) + f(y^3)$ $ \rightarrow$ $ f(x) + f(y) = f(x + y)$ simple induction gives : $ f(x_1 + x_2 + ... + x_n) = f(x_1) + f(x_2) + f(x_3) + ... + f(x_n)$ so $ \forall \ n\in\mathbb{R}$ then $ f(nx) = nf(x)$ and thus by setting $ x = 1$ we have $ f(n) = nf(1)$, since $ f(1)$ is a constant then by assuming $ f(1) = a$ , gives $ f(n) = an$ with $ a\in\mathbb{R}$ done you only solve equation on $ N$ not $ R$ We have:$ f(x + y) = f(x) + f(y)$ and $ f(x^3) = xf(x^2)$,$ (x) = - f( - x)$,$ f(kx) = kf(x)$ for all $ k \in N$ We have: $ f((x + 1)^3 + (x - 1)^3) = (x + 1)f(x^2 + 2x + 1) + (x - 1)f(x^2 - 2x + 1) = 2xf(x^2) + 2xf(1) + 4f(x)$ and $ f((x + 1)^3 + (x - 1)^3) = f(2x^3 + 6x) = 2xf(x^2) + 6f(x)$ So $ f(x) = xf(1)$ for all $ x \in R$

Sorry there are some mistakes in my solution Let me write my full solution First from my first post, we have $ f(x_1)+f(x_2)+...+f(x_n)=f(x_1+x_2+x_3+...+x_n)$ with $ n\in\mathbb{N}$ so $ f(nx)=nf(x)$ $ \forall n\in\mathbb{N}$ and then by setting $ x=m+1$ and $ y=m-1$ it gives $ f((m+1)^3+(m-1)^3)=f((m^3+3m^3+3m+1)+(m^3-3m^3+3m-1))=f(2m^3+6m)=f(2m^3)+f(6m)=2f(m^3)+6f(m)$ and $ f((m+1)^3+(m-1)^3)=(m+1)f(m^2+2x+1)+(m-1)f(x^2-2x+1)=(m+1)\left(f(m^2)+f(2m)+f(1)\right)+(m-1)\left(f(m^2)+f(-2m)+f(1)\right)=2mf(m^2)+4f(m)+2mf(1)$ since $ 2mf(m^2)=2f(m^3)$ , so $ f((m+1)^3+(m-1)^3)=2mf(m^2)+4f(m)+2mf(1)=2f(m^3)+4f(m)+2mf(1)$ $ 2f(m^3)+6f(m)=f((m+1)^3+(m-1)^3)=2f(m^3)+4f(m)+2mf(1)$ $ 2f(m)=2mf(1)$ $ \rightarrow$ $ f(m)=mf(1)$ $ \forall m\in\mathbb{R}$ please check.... EDITED : Sorry I didn't know that my solution is similar to math10's solution

not so hard. let $y=0$,then $f(x^3)=xf(x^2)$ so $f(x^3+y^3)=f(x^3)+f(y^3)$ hence f satisfies Cauchy's function,so for rational x,$f(x)=x$ hence $f(1)=1$ by letting $x=x+1$ we get $f((x+1)^3)=(x+1)f((x+1)^2)$ hence $2f(x^2)=(2x-1)f(x)+x$ let $x=x+1$ we get $2(f(x^2)+2f(x)+1)=(2x+1)(f(x)+1)+x+1$ hence $f(x)=x$.

littletush wrote: ... hence f satisfies Cauchy's function, Right littletush wrote: so for rational x,$f(x)=x$ Wrong : $f(x)=ax$

pco wrote: littletush wrote: ... hence f satisfies Cauchy's function, Right littletush wrote: so for rational x,$f(x)=x$ Wrong : $f(x)=ax$ oh gush!such a terrible mistake! a can be any real number.

So which solution is correct?????

It is easy to show that ∃$a\in \mathbb R$ s.t. $f(x)=ax(\forall x\in \mathbb Q)$. If we show that $f$ is -continuous at some point -monotonous -either upperbounded or lowerbounded on some open interval $f(x)=ax(\forall x\in \mathbb R)$ Anyone has idea?

Raja Oktovin wrote: Find all functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying \[ f(x^3+y^3)=xf(x^2)+yf(y^2)\]for all real numbers $ x$ and $ y$. Hery Susanto, Malang Let $P(x,y)$ be the assertion $f(x^3+y^3)=xf(x^2)+yf(y^2)$ Let $a=f(1)$ $P(0,0)$ $\implies$ $f(0)=0$ $P(x,0)$ $\implies$ $f(x^3)=xf(x^2)$ And so $f(x^3+y^3)=f(x^3)+f(y^3)$ and so $f(x)$ is additive. So $f(px)=pf(x)$ $\forall x$ and $\forall p\in\mathbb Q$ Let $x\in\mathbb R$ and $k\in\mathbb Q$ $P(x+k,0)$ $\implies$ $f(x^3+3kx^2+3k^2x+k^3)=(x+k)f(x^2+2kx+k^2)$ Which may be written $k^2(f(x)-ax)+2k(f(x^2)-xf(x))=0$ Considering this as a polynomial in $k$ with infinitely many roots (any rational), we get thet it must be the zero polynomial and so, looking at coefficient of $k^2$ : $\boxed{f(x)=ax\text{ }\forall x}$ which indeed is a solution, whatever is $a\in\mathbb R$

My solution:Plug x=y=0 so f(0)=0. Plug x=0 so f(x^3)=xf(x^2).(1) So f(x^3+y^3)=f(x^3)+f(y^3) for every x;y real. So f is addictive. Now we will caculate f( (x+1)^3+(x-1)^3) in 2 ways. f( (x+1)^3+(x-1)^3) = (x+1)f((x+1)^2)+(x-1)f((x-1)^2). =(x+1)[f(x^2)+2f(x)+f(1)]+(x-1)(f(x^2)-2f(x)+f(1)]. =2xf(x^2)+2xf(1)+4f(x).(2) On the other hand, f( (x+1)^3+(x-1)^3) =f(2x^3+6x)=2f(x^3)+6f(x).(3) By (1);(2) and (3) we have : f(x)=cx(c=f(1)).

@pco,@anhtaitran What a nice solution! Are you a FE master?

pco wrote: And so $f(x^3+y^3)=f(x^3)+f(y^3)$ and so $f(x)$ is additive. So $f(px)=pf(x)$ $\forall x$ and $\forall p\in\mathbb Q$ How do I conclude this? Induction would work on \(\mathbb N \) but what can I do here?

Evenprime123 wrote: pco wrote: And so $f(x^3+y^3)=f(x^3)+f(y^3)$ and so $f(x)$ is additive. So $f(px)=pf(x)$ $\forall x$ and $\forall p\in\mathbb Q$ How do I conclude this? Induction would work on \(\mathbb N \) but what can I do here? You should consider this as a well-known property of additive functions. If you want to prove it : $f(2x)=f(x)+f(x)=2f(x)$ $f(3x)=f(2x)+f(x)=3f(x)$ And, with induction : $f(nx)=nf(x)$ $\forall x$, $\forall n\in\mathbb N$ So $f(x)=f(q\frac xq)=qf(\frac xq)$ and so $f(\frac xq)=\frac 1qf(x)$ $\forall x$, $\forall q\in\mathbb N$ So $f(\frac pqx)=pf(\frac xq)=\frac pqf(x)$ and so $f(px)=pf(x)$ $\forall x$, $\forall p\in\mathbb Q^+$ It remains to remember that $f(-x)=-f(x)$ and $f(0)=0$ and so $f(px)=pf(x)$ $\forall x$, $\forall p\in\mathbb Q$

GeronimoStilton wrote: Some progress: ... ... so I can't finish the problem. Sorry, but what is the interest of this post ? Just publicly claim that you are working on this problem ? And what about your progress about your sister's garden ? and what is the last film you liked ? Just request for some help ? : read the thread ! there is already a full solution (see post #10) Just bring some new informations to the community ? but this contribution already has been given at least thrice in the current thread (and btw is the beginning of the full solution in the post #10)

This is a folklore: $\bigstar \color{blue}{\textit{\textbf{ANS:}}}$ $f(x)=cx \quad x\in \mathbb{R}$ where $c\in \mathbb{R}$. $\spadesuit \color{red}{\textit{\textbf{Proof:}}}$ It's easy to see that this is a solution to the given FE. Let $P(x,y)$ denote the given assertion, \[P(x,-x): f(0)=0\]\[P(x,0): f(x^3)=xf(x^2).\]So, let $u=x^3, v=y^3$, $f(u+v)=f(x^3+y^3)=xf(x^2)+yf(y^2)=f(x^3)+f(y^3)=f(u)+f(v)$ and $f$ is additive. We use a standard trick: \[P(x+1,x-1): \boxed{f(x)=f(1)x \quad \forall x\in \mathbb{R}}\]as many terms cancel nicely due to additivity and it yields the desired result. $\quad \blacksquare$

Raja Oktovin wrote: Find all functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying \[ f(x^3+y^3)=xf(x^2)+yf(y^2)\]for all real numbers $ x$ and $ y$. Hery Susanto, Malang $\color{blue} \boxed{\textbf{Answer: f(x)=cx}}$ $\color{blue} \boxed{\textbf{Proof:}}$ $\color{blue} \rule{24cm}{0.3pt}$ $$f(x^3+y^3)=xf(x^2)+yf(y^2)...(\alpha)$$In $(\alpha) y=0:$ $$\Rightarrow f(x^3)=xf(x^2)...(\beta)$$$(\beta)$ in $(\alpha):$ $$\Rightarrow f(x^3+y^3)=f(x^3)+f(y^3)$$$$\Rightarrow f(x+y)=f(x)+f(y)...(I)$$In $(\beta) x=x+1:$ $$\Rightarrow f(x^3+3x^2+3x+1)=(x+1)f(x^2+2x+1)$$By $(I):$ $$\Rightarrow f(x^3)+f(3x^2)+f(3x)+f(1)=xf(x^2)+xf(2x)+xf(1)+f(x^2)+f(2x)+f(1)$$$$\Rightarrow f(x)=xf(1)$$$$\Rightarrow \boxed{\textbf{f(x)=cx}}_\blacksquare$$$\color{blue} \rule{24cm}{0.3pt}$

This problem is very similar to Macedonian Mathematical Olympiad 2007 P4

F10tothepowerof34 wrote: This problem is very similar to Macedonian Mathematical Olympiad 2007 P4 It's the same problem...