Let $P(x,y)$ denote the original proposition. Clearly the only constant solution is $$\boxed{f(x)=0 \ \ \forall x \in \mathbb{R}}$$so assume $f$ is non-constant from now on. $P(x,0)$ $\implies$ $f(xf(0))=f(0)^3$ for all $x$ $\implies$ $f(0)=0$ since $f$ is non-constant. Claim 1: $f(y^3)=f(y)^3$ for all $y$. Proof: $P(0,y)$. $\blacksquare$ Claim 2: $f$ is injective at $0$. Proof: Suppose $f(a)=0$ for some $a \neq 0$. Then $P(x,a)$ $\implies$ $af(x)=f(a^3)$ $\implies$ $f$ is constant, contradiction! Claim 3: $f(1)=1$ Proof: Putting $y=1$ in Claim 1 gives $f(1)=0$, $-1$ or $1$. $0$ can be ruled out by Claim 2. If $f(1)=-1$, then $P(x,1)$ $\implies$ $f(1-x)=f(x)-1$ for all $x$. $$\implies f(x)=f(1-(1-x))=f(1-x)-1=f(x)-2$$which is absurd. Therefore $f(1)=1$. $\blacksquare$ Claim 4: $f(x+1)=f(x)+1$ for all $x$. Proof: $P(x,1)$. $\blacksquare$ Now for any $y,z$ with $y \neq 0$, we can choose an $x$ such that $z=xf(y)+y^3$ by Claim 2. Then $P(x+1,y)$ gives \begin{align*} f(xf(y)+y^3+f(y)) &=yf(x+1)+f(y)^3 \\ \implies f(z+f(y)) &= yf(x)+f(y)^3+y \ \ \dots \ \ \text{from Claim 4} \\ &= f(xf(y)+y^3)+y \ \ \dots \ \ \text{from } P(x,y) \\ \implies f(z+f(y)) &= f(z)+y \end{align*}Clearly the above holds when $y=0$ as well, so call it $Q(z,y)$. Claim 5: $f(f(y))=y$ for all $y$. Proof: $Q(0,y)$. $\blacksquare$ Now $Q(z,f(y))$ along with Claim 5 $\implies$ $f(y+z)=f(y)+f(z)$ for all $y,z$ $\implies$ $f$ is additive. Using the additivity in $P(x,y)$, we get $$f(xf(y))+f(y^3)=yf(x)+f(y)^3$$$$\implies f(xf(y))=yf(x)$$by Claim 1. Replacing $y$ by $f(y)$ in the above and using Claim 5, we get $f(xy)=f(x)f(y)$ for all $x,y$ $\implies$ $f$ is multiplicative as well. It is well known that the only non-constant function which is both additive and multiplicative is $$\boxed{f(x)=x \ \ \forall x \in \mathbb{R}}$$$\blacksquare$

Ans. $f(x) = 0 \ \ \forall x \in \mathbb{R}$ or $f(x) = x \ \ \forall x \in \mathbb{R}$. It's easy to check that it does satisfies the condition, we now prove the uniqueness. Let $P(x, y)$ denote the original proposition. $$P(x, 0) \rightarrow f(xf(0)) = f(0)^3 \implies f(0) = 0, \text{or $f$ is a constant.} $$$$P(x, 1) \rightarrow f(xf(1)+1) = f(x) + f(1)^3$$If $f(1) \neq 1$, let $x = \frac{1}{1-f(1)}$, then $f(1) = 0$, which implies $f(x) = 0$. So consider that $f(1) = 1$, we now have$$f(x+1) = f(x) +1$$Compares $P(x+1, y)$ and $P(x, y)$, we have $$f(xf(y) + y^3 +f(y)) - f(xf(y) + y^3) = y$$Noticed that $f(y) \neq 0$ when $y\neq 0$, and when $x$ takes over all real number, $xf(y) + y^3$ takes over all real number. So even when $y = 0$, we have $$f(x+f(y)) = f(x) +y$$Let $x=0$ we get $f(f(y)) = y$, replaces $y$ by $f(y)$ we have $$f(x+y) = f(x) +f(y)$$Ans also $$ yf(x) +f(y)^3 = f(xf(y) +y^3) = f(y^3) +f(xf(y)) \implies yf(x) = f(xf(y))$$Replaces $y$ by $f(y)$ we get $$f(xy) = f(x)f(y)$$Hence $f$ is multiplicative and additive, which concludes that $f(x) = x$, we're done. $\blacksquare$

Let $P(x,y):f(xf(y)+y^3)=yf(x)+f(y)^3$. (case1) $f$ is constant the only answer is $$f(x)=0$$ case2) $f$ is not constant $P(x,0): f(xf(0))=f(0)^3\rightarrow f(0)=0$. claim1) $f(x)=0<->x=0$ <pf> say that there exists such $t \neq 0$ that satisfies $f(t)=0$. $P(x,t): f(t^3)=tf(x) \rightarrow f$ is constant which is a contradiction. claim2) $f(1)=1$ <pf> $P(0,1): f(1)=f(1)^3 \rightarrow f(1)=1 ,-1$ ( $f(1) \neq 0$ since claim 2) if $f(1)=-1$, then $P(x,1): f(-x+1)=f(x)-1$. So, $f(-x+1)=f(x)-1=f(-x+1)-2$ which is a contradiction. So, $f(1)=1$. claim3) $f(x+y)=f(x)+f(y)$ <pf> since $f(1) = 1$, $P(x,1): f(x+1) = f(x) +1$. Comparing $P(x+1, y)$ and $P(x, y) \rightarrow f(xf(y) + y^3 +f(y)) - f(xf(y) + y^3) = y$ Because of claim1), $xf(y)+y^3$ can express every real number, so $\rightarrow f(x+f(y))=f(x)+y$. when $x=0$, it means $f(f(y))=y\rightarrow f(x+y)=f(x+f(f(y)))=f(x)+f(y)$. claim4) $f(x)=x$ <pf> $P(0,y): f(y^3)=f(y)^3$. so, $f(y^3)+3f(y^2)+3f(y)+1=f((y+1)^3)=f(y+1)^3=(f(y)+1)^3=f(y)^3+3f(y)^2+3f(y)+1 \rightarrow f(y^2)=f(y)^2$. since if $x>0, f(x)=f(\sqrt{x})^2>0$. since $f$ is additive, which concludes that $f(x) = x$.

The solutions are $f(x)=x$ and $f(x)=0$. Let $P(x,y)$ denote the original assertion. From $P(0,0)$, we get that $f(0)=f(0)^3$. Thus a case bash ensues. 1.Case The first case is when $f(0) \neq 0$ This means that $f(0)^2=1$. Now a case bash ensues. 1.Subcase $f(0)=1$, from $P(x,0)$ we have that $f(x)=1$, but checking this gives us that this isn't a solution. 2.Subcase $f(0)=-1$, from $P(x,0)$ we have that $f(x)=-1$, but checking this gives us that this isn't a solution. 2.Case: The second case is when $f(0)=0$. From $P(0,x)$ we have that $f(y^3)=f(y)^3$. So now let's say that there is a set $S$ such that it contains all the numbers $x$ so that $f(x)=0$, written in notation $S=\{x | f(x)=0\}$. 1.Case: Assume that $card(S)\geq 2$. Let say that $y$ in our asserion is in $S$, but so that $y \neq 0$. This gives us that $f(y^3)=yf(x)+f(y^3)$, which implies $yf(x)=0$, but since $y$ isn't $0$, we get that $f(x)=0$ and since $x$ was arbitary we have that $f(x) =0$, for all real $x$. 2.Case: Assume that $card(S)=1$, i.e. we have that $f$ is injective at the point $x=0$. From $f(y^3)=f(y)^3$, we have that $f(1)=f(1)^3$. Now a case bash ensuses. 1.Subcase: $f(1)=1$, from $P(x,1)$ we have that: $$f(x+1)=f(x)+1$$by induction this implies that $f(x)=x$ for all $x \in \mathbb{Z}$. From $P(x,y)$, where $x\in \mathbb{Z}$, we have that $f(xf(y)+y^3)=xy+f(y^3)$ and from $P(x,y)$, when $y$ is an integer we have that $f(xy+y^3)=yf(x)+y^3$. Since we have that ($x \in \mathbb{Z}$, $x \neq 0$): $$f(xf(y)+y^3)-f(y^3)=xy=yx=f(yf(x)+x^3)-f(x^3)=f(xy+x^3)-x^3=xf(y)$$This implies that $f(y)=y$. 2.Subcase: $f(1)=-1$, from $P(x,1)$ we have that: $$f(1-x)=f(x)-1$$but notice that we must have that: $$f(-1)=f(2)-1 \; \; \text{and} \; \; f(2)=f(-1)-1$$to hold simultaneously, but that isn't possible, since we get the following situation $f(2)=f(2)-1-1 \implies 0 \neq -2$. Thus we have no solution in this subcase.

Another way of finishing the problem once you get Cauchy's FE is this: https://artofproblemsolving.com/community/c6h545068p3151936 (the exact same argument works with 3 instead 2013)

mathematics2003 wrote: Determine all $ f:R\rightarrow R $ such that $$ f(xf(y)+y^3)=yf(x)+f(y)^3 $$ Here is a solution with another way of finding that f is an involution. $P(x,0) \implies f(xf(0))=f(0)^3$ So, if $f(0)\neq 0$, then $f$ is just a non-zero constant, which fails. So, $f(0)=0$ and $P(0,y)\implies f(y^3)=f(y)^3$ Note that all-zero works. Henceforth assume that $\exists u$ such that $f(u)\neq 0$. If $r$ is a root, $P(u,r)\implies r=0$. So, $f(1)\neq 0$ and if $f(1)\neq 1$, $P(\frac{1}{1-f(1)},1)$ gives a contradiction. Hence, $f(1)=1$. $P(x,1)\implies f(x+1)=f(x)+1$. Let $y\neq 0$. Then, $P(\frac{-y^3}{f(y)},y)\implies yf(\frac{-y^3}{f(y)})=-f(y)^3$ And $P(\frac{f(y)-y^3}{f(y)},y)\implies f(f(y))=yf(\frac{f(y)-y^3}{f(y)})+f(y)^3=yf(1+\frac{-y^3}{f(y)})+f(y)^3=y+\frac{-y^3}{f(y)}+f(y)^3=y$. Also, $f(f(y))=y$ holds for $y=0$ clearly. So, $f$ is an involution. If $y\neq 0$, then $P(x+1,y)\implies f(xf(y)+f(y)+y^3)=yf(x)+y+f(y)^3=f(xf(y)+y^3)+y$. So, $f(z+f(y))=f(z)+y$ and using involution: $$f(z+y)=f(z)+f(y)$$Note that this obviously hold even if $y=0$, so $f$ is additive. Now, $yf(x)+f(y)^3=f(xf(y)+y^3)=f(xf(y))+f(y^3)\implies f(xf(y))=yf(x)\implies f(xy)=f(x)f(y)\implies \forall a\geq 0, f(a)\geq 0$, which with additive implies $f(x)=x, \forall x$ Hence, the only solutions are $f(x)=x, \forall x$ and $f(x)=0, \forall x$, which both fit.

I liked it Let $P(x,y)$ be the given assertion. $\boxed{f(x)=0}$ works, and is the only constant solution. Now assume $\exists j:f(j)\ne0$, true iff $f$ is nonconstant. $\textbf{1. }f(0)=0$ If $f(0)\ne0$ then: $P\left(\frac x{f(0)},0\right)\Rightarrow f(x)=f(0)^3$, contradiction. $\textbf{2. }f\left(x^3\right)=f(x)^3$ $P(0,x)$. $\textbf{3. }f(1)=1$ $P(0,1)\Rightarrow f(1)=f(1)^3\Rightarrow f(1)\in\{-1,0,1\}$ If $f(1)=0$ then $P(x,1)\Rightarrow f(x)=0$, contradiction. If $f(1)=-1$ then $P(x,1)\Rightarrow f(1-x)=f(x)-1$, but $P(1-x,1)\Rightarrow f(x)=f(1-x)-1$, contradiction. $\textbf{4. }f(k)=0\Rightarrow k=0$ (injectivity at $0$) $P(j,k)\Rightarrow kf(j)=f\left(k^3\right)=f(k)^3=0\Rightarrow k=0$ $\textbf{5. }Q(x,y):f(x+f(y))=y+f(x)$ $P(x,1)\Rightarrow f(x+1)=f(x)+1$ $P\left(\frac{x-y^3}{f(y)}+1,y\right)-P\left(\frac{x-y^3}{f(y)},y\right)\Rightarrow f(x+f(y))=y+f(x)$ and since this also holds for $y=0$, we are done. $\blacksquare$ $Q(0,y)\Rightarrow f(f(y))=y$ $Q(x,f(y))\Rightarrow f(x+y)=f(x)+f(y)$ Using this in the original assertion, we obtain $R(x,y):f(xf(y))=yf(x)$ $R(x,f(y))\Rightarrow f(xy)=f(x)f(y)$ and it's well-known that the only function that is both additive and multiplicative is the identity function, $\boxed{f(x)=x}$.

mathematics2003 wrote: Determine all $ f:R\rightarrow R $ such that $$ f(xf(y)+y^3)=yf(x)+f(y)^3 $$ Nice Problem. I enjoyed it a lot. My idea is also same as some of the above ones. Claim 1-: $f(x)\neq 1$ for all $x\in R$ Prove-: FTSOC $f(x)=1$ for any $x\in R$ But then $f(x+y^3)=y+1$ will be the contradiction. Claim 2-: $f(x)\neq -1$ for all $x\in R$ Prove -: FTSOC $f(x)=-1$ But then $f(-x+y^3)=-y+1$ will be the contradiction. Claim 3-: $f(0)=0$ Prove-: let $P(0, 0)\implies f(0)=f(0)^3$ Hence $f(0)=0$ or $1$ or $-1$ Case 1-: if $f(0)=1$ then $P(x, 0)\implies f(x)=f(0)^3=1$ for any $x\in R$ Which is not possible by claim $1$ Case 2-: if $f(0)=-1$ then again $P(-x, 0)\implies f(x)=-1$ Which is not possible by claim $2$ Now as both cases above are not possible so $f(0)=0$ Claim 4-: $f(x^3)=f(x)^3$ for any $x\in R$ Prove -: $P(0, x)\implies f(x^3)=xf(0)+f(x)^3$ Claim 5-: if $f$ is a constant function then $f(x)=0$ is the only possibility. Prove -: let's say $f(x)=k$ for some constant $k$ then by Claim $4$ we get $k=k^3\implies k=1$ or $-1$ or $0$ Now By Claim $1,2$ we get $f(x)=0$ is only possibility if $f$ is constant. Claim 6-: $f(1)=1$ Prove-: By claim $4$ we have either $f(1)=0$ or $-1$ or $1$. Case 1-: $f(1)=0$ Then $P(x, 1)\implies f(1)=f(x)+f(1)^3\implies f(x)=0$ Case 2-: $f(1)=-1$ Then $P(1, 1)\implies f(0)=f(1)-1=-2$ which is contradiction. So $f(1)=1$ Claim 7-: $f(x+1) =f(x)+1$ Prove -: $P(x, 1)\implies f(x+1)=f(x)+1$ Claim 8-: We can't have any other $x_1\in R-\{0\}$ such that $f(0)=f(x_1)=0$ for Non constant function $f$ Prove-: $P(x, x_1)\implies f((x_1)^3)=x_1f(x)+0^3$ By claim $4$ $f((x_1)^3)=0$ Hence we get $f(x)=0$ for all $x\in R$ But this is a contradiction as $f$ is Non constant. Claim 9-:$f$ is an involution Prove-: $P(x+1, y)\implies f((x+1)f(y)+y^3)=yf(x+1) +f(y)^3$ By claim $7$ we have $f(xf(y)+y^3+f(y))=f(xf(y)+y^3)+y$ Now we can choose $xf(y)+y^3=0$ then By claim $8$ We can get $f(f(y))=y$ Claim 10-: $f(x+y)=f(x)+f(y)$ Prove -: which is just Direct by Claim $9$ as $xf(y)+y^3, f(y) $ can cover every real Number. So we can rewrite it as $f(xf(y)+y^3+f(y))=f(xf(y)+y^3)+f(f(y))$ Claim 11-: $f(x)=x$ is the only Non constant Solution for $f$ Prove-: By Claim $10$ we have $f(xf(y)+y^3)=f(xf(y))+f(y^3)=yf(x)+f(y)^3\implies f(xf(y))=yf(x)=f(f(y))f(x)\implies f(xy)=f(x)f(y)$ Hence $f(x)=x$ $\blacksquare$

A slightly different way to show that $f$ is additive: We quickly obtain that $f(x) = 0 \Leftrightarrow x = 0$ and $f(x^3) = f(x)^3$. We need to first show that $f(1) = 1$. Indeed, by $f(x^3) = f(x)^3 \Rightarrow f(1)^2 = f(-1)^2 = 1 \Rightarrow f(1), f(-1) = \pm 1$. On the other hand, by injectivity at $0$ we have $f\left(\frac{-x^3}{f(x)} \right) = \frac{-f(x)^3}{x}$. Hence if $f(1) = -1$ then $f\left(\frac{-x^3}{f(x)} \right) = \frac{-f(x)^3}{x} \Rightarrow f(1) = 1$, contradiction! Similarly, we obtain that $f(-1) = -1$. Now, $P(x,1)$ gives $f(x+1) = f(x)+1 \Rightarrow f(x+n) = f(x) + n \forall n \in \mathbb{Z}$ by induction and $P(x,-1)$ gives $f(-x-1) = -f(x)-1 = -f(x+1)$, hence $f$ is an odd function. $P(x^{1/3},y/f(x^{1/3}))+P(x^{1/3},-y/f(x^{1/3})): f(x+y) + f(x-y) = 2f(x)$. $P(x,2): f(xf(2) + 8) = 2f(x) + f(2)^3$. From $f(x+n) = f(x) + n$ and $f(2) = f(1) + 1 = 2$ we have $f(2x) = 2f(x)$. Hence, $f(a)+f(b) = f(a+b)$ where $a = x+y, b = x-y$. This allows us to obtain $f(xf(y)) = yf(x) \Rightarrow f(f(y)) = y \forall y \Rightarrow f(xy)=f(x)f(y) \forall x,y \Rightarrow f(x) = xf(1) = x \forall x$ since multiplicativity implies $f$ is increasing, which also implies $f$ is linear since it is additive. Hence proved.

mathematics2003 wrote: Determine all $ f:R\rightarrow R $ such that $$ f(xf(y)+y^3)=yf(x)+f(y)^3 $$ Easy for a P5. Let \(P(x,y)\) denote the assertion that \(f(xf(y)+y^3)=yf(x)+f(y)^3\). We begin by proving some claims. Claim 1. \(f(x+1)=f(x)+1\) for all reals \(x\). Proof. Firstly, we will evaluate \(f(0)\). \(P(0,0)\) gives us that \(f(0)=f(0)^3\), so \(f(0)=0,\pm1\). Assume that \(f(0)=\pm1\). Then, \(P(x,0)\) gives us that \[f(\pm x)=\pm1\]for all reals \(x\), implying that \(f\) is constant and that too equal to \(\pm1\), a contradiction, therefore \(f(0)=0\). Now, \(P(0,1)\) gives us that \(f(1)=f(1)^3\), so \(f(1)=0,\pm1\). If \(f(1)=0\), then \(P(x,1)\) gives us that \(f(x)=0\) for all reals \(x\), which is indeed a solution. If \(f(1)=-1\), then \(P(x,1)\) gives us that \[f(1-x)=f(x)-1\]and \(P(1-x,1)\) gives us that \(f(x)=f(1-x)-1\), a contradiction. Therefore, \(f(1)=1\). Now, \(P(x,1)\) gives us that \(f(x+1)=f(x)+1\) for all reals \(x\), as claimed. $\blacksquare$ Claim 2. \(f\) is involutive. Proof. Comparing \(P(x,y)\) and \(P(x+1,y)\) and using Claim 1, we see that \[f(xf(y)+y^3+f(y))=(yf(x)+f(y)^3)+y\]or \[f(xf(y)+y^3+f(y))-f(xf(y)+y^3)=y\]Plugging \(x=\frac{-y^3}{f(y)}\) in this, we get \(f(f(y))=y\) for all reals \(y\), as desired. $\blacksquare$ Claim 3. \(f\) is additive. Proof. From claim 2, it is clear the \(f\) is bijective. Now, since \(f(y)\) and \(xf(y)+y^3\) span all the real numbers, and we have that \[f(xf(y)+y^3+f(y))-f(xf(y)+y^3)=y=f(f(y),\]we conclude that \(f\) is additive, as desired. $\blacksquare$ Now, since \(f\) is additive and \(f(x^3)=f(x)^3\) for all reals \(x\), by generalized Calvin Deng's ELMO 2013 SL, we conclude that \(f(x)=x\) for all reals \(x\). $\blacksquare$ Remark. If you replace \(3\) with any odd positive integer \(n\), the same procedure follows.

Solved with NoOne becuase i can do it myself lol As always let $P(x,y)$ the assertion of this F.E. and assume that $f$ is not constant as $f \equiv 0$ works. Claim 1: $f(x^3)=f(x)^3$ Proof: $P(x,0)$ $$f(xf(0))=f(0)^3 \implies f(0)=0$$$P(0,x)$ $$f(x^3)=f(x)^3$$Claim 2: $f(x+1)=f(x)+1$ Proof: Assume that $f(1) \ne 1$, then by $P \left(\frac{1}{1-f(1)},1 \right)$ $$f(1)=0$$And now by $P(x,1)$ $$f(x)=0 \; \text{contradiction!!}$$Now by $P(x,1)$ $$f(x+1)=f(x)+1$$Claim 3: $f$ bijective Proof: $P \left(\frac{1}{f(x)}, x \right)$ $$f \left(\frac{1}{f(x)} \right)=\frac{1}{x} \implies f \; \text{bijective}$$Claim 4: $f$ involutive and additive. Proof: $P \left(1-\frac{x^3}{f(x)}, x \right)$ for $x \ne 0$ $$f(f(x))=xf \left(1-\frac{x^3}{f(x)} \right)+f(x)^3=x \implies f \; \text{involution}$$$P \left(1-\frac{x-f(y)^3}{y}, f(y) \right)$ for $y \ne 0$ $$f(x+y)=f(y)f \left(1-\frac{x-f(y)^3}{y} \right)+y^3=f(x)+f(y) \implies f \; \text{additive}$$Finishing: Using Claim 4 with $P(x,f(y))$ $$f(xy)=f(x)f(y) \implies f \; \text{multiplicative} \; \implies f(x)=x$$Thus we are done

Let $P(x,y)$ denote given assertion. Clearly the only constant solution is $\boxed{f\equiv 0}$, so henceforth assume $f$ isn't constant. $P(x,0): f(xf(0))=f(0)^3$. If $f(0)\ne0$, then $f$ is constant. So $f(0)=0$. $P(0,x): f(x^3)=f(x)^3$. Claim: If $f(x)=0$, then $x=0$. Suppose some $k\ne0$ satisfies $f(k)=0$. $P(0,k): f(k^3)=0$. $P(x,k): f(k^3)=kf(x)=0$. This implies $f\equiv 0$, a contradiction. $\blacksquare$ Note that $f(1)=f(1)^3\implies f(1)\in \{-1,0,1\}$. Claim: $f(1)=1$. Clearly, $f(1)\ne0$. So we will prove $f(1)\ne -1$. Suppose $f(1)=-1$. $P(x,1): f(1-x)=f(x)-1$. Setting $x=\frac{1}{2}$ here gives $f\left(\frac{1}{2}\right)=f\left(\frac{1}{2}\right)-1$, a contradiction. $\blacksquare$ $P(x,1): f(x+1)=f(x)+1$. For $x\ne0$, \[P\left(\frac{1}{f(x)},x\right): f(x^3+1)=f(x)^3+1=xf\left(\frac{1}{f(x)}\right)+f(x)^3\implies xf\left(\frac{1}{f(x)}\right)=1\implies f\left(\frac{1}{f(x)}\right)=\frac{1}{x}\] Claim: $f$ is bijective. Proof: Suppose $f(a)=f(b)$. Then $f\left(\frac{1}{f(a)}\right)=\frac{1}{a}$ and $f\left(\frac{1}{f(a)}\right)=\frac{1}{b}$. Thus, $a=b$ and $f$ is injective. Since $f(0)=0$, $f$ is surjective. $\blacksquare$ Claim: $f$ is involutive. $P(x+1,y): f(xf(y)+f(y)+y^3)=yf(x)+y+f(y)^3=f(xf(y)+y^3)+y$ Consider $x=\frac{-y^3}{f(y)}$. Then we get $f(f(y))=y$. $\blacksquare$ Claim: $f$ is additive. From $f$ bijective, it's clear that $f(y)$ and $xf(y)+y^3$ span across all reals. We have \[f((xf(y)+y^3)+f(y))=f(xf(y)+y^3)+y=f(xf(y)+y^3)+f(f(y)),\]so $f$ is additive. $\blacksquare$. The finish: Note that $f(x+1)^3=(f(x)+1)^3=f(x)^3+3f(x)^2+3f(x)+1$. We will use $f(3x)=f(x+x+x)=3f(x)$ in the following step: \[f(x+1)^3=f(x^3+3x^2+3x+1)=f(x^3)+f(3x^2)+f(3x)+1=f(x)^3+3f(x^2)+3f(x)+1.\] So $f(x)^3+3f(x)^2+3f(x)+1=f(x)^3+3f(x^2)+3f(x)+1\implies f(x^2)=f(x)^2$. Thus, $f$ is bounded over the interval $(0,\infty)$. So $f(x)=kx$ for some $k$. Checking gives $k\in \{-1,0,1\}$. It's obvious $k=-1$ and $k=0$ don't work, so $\boxed{f(x)=x}$.

The identically vanishing function works. So WLOG $f(x)\neq 0$ for some $x.$ Denote the given assertion by $P(x,y).$ $P(x,0)$ forces $f(0)=0.$ Moreover $P(0,x)$ gives $f(x^3)=f(x)^3.$ If $f(1)\neq 1$ then $P((1-f(1))^{-1},1)$ implies $f(1)=0$ which according to $P(x,1)$ implies $f$ is constant, absurd. So $f(1)=1$ whence $f(x+1)=f(x)+1.$ In particular $P(f(x)^{-1},x)$ yields surjectivity. Compare $P(x+1,y)$ with $P(x,y)$ (using surjectivity) to get additivity. But then according to $P(x,y)$ we deduce $f$ is also multiplicative. So $f$ is the identity, which works.

$f(xf(y)+y^3)=yf(x)+f(y)^3$ $P(0,0)$ $f(0)=f(0)^3$ $f(0)=0,1,-1$ Case1: $f(0)=1$ $P(x,0)$ $f(x)=f(0)=1$ contradicition. Case2: $f(0)=-1$ $P(x,0)$ $f(-x)=f(0)=-1$ contradicition. So, $f(0)=0$ $P(0,x)$ $f(x^3)=f(x)^3$ $P(0,1)$ $f(1)=0,-1,1$ Case1: $f(1)=0$ $P(x,1)$ $f(1)=f(x)=0$ which is a solution Case2: $f(1)=-1$ $P(1,1)$ $f(0)=2$ which is contradicition. Case3: $f(1)=1$ $P(x,1)$ $f(x+1)=f(x)+1$ $P(x+1,y)$ $f(xf(y)+f(y)+y^3)=yf(x)+y+f(y^3)=f(xf(y)+y^3)+y$ That motivates us to pick $xf(y)+y^3=p$ $f(p+f(y))=f(p)+y$ $P(0,y)$ $ff(y)=y$ $P(p,f(y))$ $f(p+y)=f(p)+f(y)$ now back to original form $f(xf(y)+y^3)=xf(y)+f(y)^3=f(xf(y))+f(y)^3$ $f(xf(y))=f(xf(y))$ $P(x,f(y))$ $f(xy)=f(x)f(y)$ $f$ multiplicative and additive so that $f(x)=x$ and we are done $f(x)=x,0$ for all x

Nice Problem! The solutions are $f \equiv 0,x$ which clearly work, WLOG $f$ be non constant. Claim 1: $f(0)=0$

Proof

Claim 2: $f(y)^3=f(y^3)$

Proof

Claim 3: $f$ is injective at $0$

Proof

Claim 4: $f(1)=1$

Proof

Claim 5:$f(x+1)=f(x)+1$

Proof

Claim 6:$f$ is an involution

Proof

Claim 7:$f$ is additive

Proof

Claim 8:$f$ is multiplicative

Proof

Hence we're done.

Easy but interesting problem, finish in the break time of midterm exam. Here’s my solution: mathematics2003 wrote: Determine all $ f:R\rightarrow R $ such that $$ f(xf(y)+y^3)=yf(x)+f(y)^3 $$ Let $P(x,y)$ be the assertion For starters, $P(0,0)$ implies $f(0)=1,-1,0$, for the first and second cases, $P(x,0)$ gives contradiction, so $f(0)=0$ Also, $P(0,y)$ implies $f(y^3)=f(y)^3$ 先煉靈丹, Suppose there exist $a\neq 0$ such that $f(a)=0$, $P(x,a)$ and we found $\boxed{f(x)=0\quad \forall x\in\mathbb{R}}$ Otherwise, $f$ is injective at $0$ $P(1,1)$ we know $f(1)=1,-1,0$, but $0$ is bad because the f is injective at $0$, also $-1$ is also bad because $P(1,1)$ will break it. So only $f(1)=1$ is possible $P(x,1)$ implies $f(x+1)=f(x)+1$ For $y\neq 0$ $P(x+1,y)$ implies $f(f(y)+xf(y)+y^3)=yf(x)+y+f(y)^3=f(xf(y)+y^3)+y$ Because $f(y)\neq 0$, notice that $x\to xf(y)+y^3$ is surjective, we can let $xf(y)+y^3=z$ As $x$ can be any real number, $z$ can also be any real number So $f(f(y)+z)=f(z)+y$ for all $y,z$ put $z=0$ we can get $f(f(y))=y$ put $y=f(z)$ we get Cauchy!! since we also know $f(x^3)=f(x)^3$, so $f$ is linear Thus, $\boxed{ f(x)=x \quad\forall x\in\mathbb{R}}$