lol denote $P(x,y)$ what it usually means. Now clearly $f$ is surjective so let $f(u)=0$. Then $P(x,u)$ gives $f(x^3)=x^2f(x)+u$. Now $x=1$ gives $u=0$, so $f(0)=0$. Now $P(x,0)$ gives $f(x^3)=x^2f(x)$. $P(0,y)$ gives $f(f(y))=y$. Therefore, $f(x^3+f(y))=x^2f(x)+y=f(x^3)+f(f(y))$. Since $f(x)$ and $x^3$ are both surjective, we deduce that $f$ is additive. Now $$f((x+1)^3+(x-1)^3)=f(2x^3+6x)=2f(x^3)+6f(x)=2x^2f(x)+6f(x)$$$$f((x+1)^3+(x-1)^3))=f((x+1)^3)+f((x-1)^3)=(x+1)^2f(x+1)+(x-1)^2f(x-1)= (x+1)^2(f(x)+f(1)) + (x-1)^2(f(x)-f(1)) = (2x^2+2)f(x) + 4xf(1)$$ These give us $f(x)=xf(1)$. Now plugging this in gives us $f(1)(x^3+f(1)y) = x^2 \cdot f(1)x + y$, so $f(1)^2=1$. Therefore, we have $f(1) = \pm 1$. This gives us $f(x) \equiv x$ and $f(x) \equiv -x $.

rkm0959 wrote: $$(x+1)^2f(x+1)+(x-1)^2f(x-1)= (x+1)^2(f(x)+f(1)) + (x-1)^2(f(x)-f(1))$$ How did you get $f(x-1) = f(x) - f(1)$. Isn't it $f(x-1) = f(x) + f(-1)$. Shouldn't we show $f$ is odd for what you did??

That's because $f(-y)+f(y)+f(y)=f(0)+f(y)=f(y)$ and so $f(-y)=-f(y)$.

1) P(0,y) ==> f(f(y))=y; f- biektiv 2) f(0)=k; P(k,y) ==> f(k^3+f(y))=y=f(f(y))==> k=0; 3) P(x,0)==> f(x^3)=×*×f(x) and f(f(y))=y ==> f(x^3+f(y))=f(x^3)+f(f(y)) f surjectiv ==> a=x^3, b=f(y); f(a+b)=f(a)+f(b); It's Koshi 4) f((x+y)^3)=(x+y)^2f(x+y) f((x+y)^3)=f(x^3)+f(3x^2y)+f(3y^2x)+f(y^3) ==> f(x)=×; and f(x)=-x

Once we prove that $f$ is additive over reals and also injective , it is enough to prove that $f(x)=cx$ for some constant $c$ right ?? Or am I missing something We then substitute to get $c= 1$ or $c=-1$

probably similiar to the others but still $P(0,x) \implies f(f(x))=x$ for all $x\in \mathbb{R}$ which implies $f$ is surjective and injective.$\square$. $P(x,0) \implies f(x^3+f(0))=x^2f(x)$ now we have $f(f(x^3+f(0)))=x^3+f(0)=f(x^2f(x))$ now letting $x=1$ we have $f(f(1))=1+f(0) \implies f(0)=0$ so we have $f(x^3)=x^2f(x)$ now we check get that $f$ is linear by $P(x,f(y^3))$ and so $f(x)=cx$ checking we get $c=1$ or $-1$ and so we are done $\blacksquare$. oh yeah @below but f is involutive and f is additive implies f is linear

@above f being additive doesn't directly imply f linear in R to R I don't think so

Meh FE, standard with the exception of the last step which gives a little insight into the method to attack such FEs- $P(0,y): f(f(y))=y$ $=>f$ is bijective $P(1,0): f(0)=0$ as $f$ is injective. $P(x,0): f(x^3)=x^2f(x)$- thus $P(x,f(y))=>f(x^3+y)=x^2f(x)+f(y)=f(x^3)+f(y)=>f(x+y)=f(x)+f(y)$ (hence $f(-x)=-f(x)$) Thus we have the 2 equations $f(x^3)=x^2f(x)$ and $f(x+y)=f(x)+f(y)$ Here comes arguably the only non-intuitive part of the problem- usually here, the drill when given Cauchy and $f(x^2)=xf(x)$ is to put in $f(x^2+2x+1)$ and then simplifying using $f((x+1)^2)=(x+1)f(x+1)$ and additivity- perhaps we can now extrapolate (although directly won't work) to cubes. A little bit of trial and error now gets you to the key observation- we want to get rid of any $f(x^2)$ term, as we've no clue of its relation to $f(x)$- thus it makes sense to substitute $(x+\text{something}^3,x-\text{something}^3)$ (I first did this with a general $x+k$, but a nice value of $k$ as $1$ works just as well)- Thus $f((x+1)^3)+(x-1)^3)=f(x+1)^3+f(x-1)^3=(x+1)^2f(x+1)+(x-1)^2f(x-1)=x^2f(x)+f(x)-2xf(x)-f(1)(x^2-2x+1)+x^2+2xf(x)+f(x)+f(1)(x^2+2x+1)$. But $f((x+1)^3)+(x-1)^3)$ also equals $f(2x^3+6x)=2x^2f(x)+6f(x)$. Hence $2x^2f(x)+6f(x)=2x^2f(x)+4f(1)x+2f(x)=>f(x)=f(1)x$. Putting $f(x)=kx$ into original equation gives $f(1)=+-1$. Thus only solutions are $f(x)=x$ and $f(x)=-x$.

$\textcolor{blue}{\text{The Answer is just }}f(x) =+ x ,-x\textcolor{blue}{\text{It's not hard to see that it works.}}$ Put $x=0$ in original equation to get $f(f(y))=y$. Now put $f(y)$ in the original equation in place of $y$ which gives \[f(x^3+f(f(y)))=x^2f(x)+f(y)\] \[\implies f(x^3+y)=x^2f(x)+f(y) \cdots (*)\] Put $x=1,y=0$ in $(*)$ to get,$\boxed{f(0)=0}$. Now put $y=0$ in $(*)$ , $f(x^3)=x^2f(x)$. From$ (*)$ get, \[f(x^3+y)=x^2f(x)+f(y)= f(x^3)+f(y)\] This is a simple Cauchy equation. So $f(x)$ will be $=cx$ . Now put $f(x)=cx$ in original equation to get , \[c(x^3+cy)=cx^3+y\] \[\implies c=+1,-1\] So we get our desired results.

ftheftics wrote: $\textcolor{blue}{\text{The Answer is just }}f(x) =+ x ,-x\textcolor{blue}{\text{It's not hard to see that it works.}}$ Put $x=0$ in original equation to get $f(f(y))=y$. Now put $f(y)$ in the original equation in place of $y$ which gives \[f(x^3+f(f(y)))=x^2f(x)+f(y)\] \[\implies f(x^3+y)=x^2f(x)+f(y) \cdots (*)\] Put $x=1,y=0$ in $(*)$ to get,$\boxed{f(0)=0}$. Now put $y=0$ in $(*)$ , $f(x^3)=x^2f(x)$. From$ (*)$ get, \[f(x^3+y)=x^2f(x)+f(y)= f(x^3)+f(y)\] This is a simple Cauchy equation. So $f(x)$ will be $=cx$ . Now put $f(x)=cx$ in original equation to get , \[c(x^3+cy)=cx^3+y\] \[\implies c=+1,-1\] So we get our desired results. This solution is incorrect/currently incomplete; as $f$ being additive does not mean that it is linear.

bora_olmez wrote: ftheftics wrote: $\textcolor{blue}{\text{The Answer is just }}f(x) =+ x ,-x\textcolor{blue}{\text{It's not hard to see that it works.}}$ Put $x=0$ in original equation to get $f(f(y))=y$. Now put $f(y)$ in the original equation in place of $y$ which gives \[f(x^3+f(f(y)))=x^2f(x)+f(y)\] \[\implies f(x^3+y)=x^2f(x)+f(y) \cdots (*)\] Put $x=1,y=0$ in $(*)$ to get,$\boxed{f(0)=0}$. Now put $y=0$ in $(*)$ , $f(x^3)=x^2f(x)$. From$ (*)$ get, \[f(x^3+y)=x^2f(x)+f(y)= f(x^3)+f(y)\] This is a simple Cauchy equation. So $f(x)$ will be $=cx$ . Now put $f(x)=cx$ in original equation to get , \[c(x^3+cy)=cx^3+y\] \[\implies c=+1,-1\] So we get our desired results. This solution is incorrect/currently incomplete; as $f$ being additive does not mean that it is linear. See for any real $a$ we must get b so that ,$a=b^3$ we can say from here $f(a+x)=f(a)+f(x)$ rest proof is same as to Cauchy functional equation . That's why I don't write the rest .

Let $P(x,y)$ denote the assertion as usual. $P(0,y) \implies f(f(y)) = y$ so $f$ is bijective. Let $u$ be such that $f(u) = 0$. $P(x,u) \implies f(x^3) = x^2f(x) + u$, plugging $x=1$ gives $u=0$, so $f(0) = 0$. So $f(x^3) = x^2f(x)$. $P(x,f(y)) \implies f(x^3 + y) = x^2f(x) + f(y) = f(x^3) + f(y)$. But as $x^3$ is surjective, we have that $f(x + y) = f(x) + f(y)$ for all real $x,y$. So using additivity we have $f((x+1)^3) = (x+1)^2f(x + 1) \implies f(x^3) + 3xf(x) + 3f(x) + f(1) = x^2f(x) + 2xf(x) + f(x) + x^2f(1) + 2xf(1) + f(1)$ And so we have $(x+2)f(x) = x^2f(1) + 2xf(1) = xf(1)(x + 2)$ and so $f(x) = xf(1)$ for all $x\neq -2$. But we can easily deduce it for $-2$ by plugging in $x=-2$ and $y=0$ in the original. So we have $f(x) = xf(1)$ for all $x$ and testing in the original equation gives us $f(1) = \pm 1$ and so we have $\boxed{f(x) = x}$ and $\boxed{f(x) = -x}$ as our solutions which clearly satisfy the given equation.

Let x=0 we have $f(f(y))=y$ implies f is bijective If x=1 and y=0 we get $f(1+f(0))=f(1)$ since f is injective $f(0)=0$ Now if y=0 we have $f(x^3)=x^2f(x)$ From this we have $f(x^3)=x^{2+2/3}f(x^{1/3})$ from induction we get $f(x^3)=x^{2+2/3+2/9+...+2/3^n}f(x^{1/3^n})$ As n approaches infinity 2+2/3+... approaches 3 and $f(x^{1/3^n}) $approaches$ f(1)$ thus $f(x^3)=x^3f(1)$ f(1) cannot be equal to 0 because of the injectivity thus $f(x)=xa$ Plugging it in to our original equation we get $a^2y=y $ thus $a=1 $ Or $a=-1 $ thus $f(x)=x $ or $f(x)=-x$ Can anyone look if this proof correct or did I assume to much up there with the limits?

Let $P(x,y)$ be the given assertion. $P(0,y)$ gives $f(f(y)) = y$ and so $f$ is bijective. Let $z$ be the unique number such that $f(z) = 0$ Then, $P(z,f(y))$ gives that $f(z^3 + y) = f(y) \implies z = 0$ by injectivity. So, $P(x,0)$ gives $f(x^3) = x^2f(x)$ and so the original assertion becomes $f(x^3 + y) = f(x^3) + f(y)$ and since $x^3$ is surjective, this means the function is additive. We get $(x+1)^2(f(x)+f(1)) = f(x+1)^3 = f(x^3 + 3x^2 + 3x + 1) = f(x^3) + 3f(x^2) + 3f(x) + f(1)$. Simplifying, we get that this means $2xf(x) + x^2f(1) + 2xf(1) = 3f(x^2) + 2f(x)$ Doing the similar thing with $(x-1)^3$, we get that $-3f(x^2) + 2f(x) = -2xf(x) - x^2f(1) + 2xf(1)$ Adding both of them, we get that $4f(x) = 4xf(1)$ and so $f(x) = cx$. Substituting in the original equation, we get that $c = \pm 1$ and so the solutions are $f(x) = x$ and $f(x) = -x$

dame dame

Solved with Rama1728 Let $P(x,y)$ the assertion of the given F.E. $P(0,x)$ $$f(f(x))=x \implies f \; \text{involution}$$$P(f(0),x)$ $$f(f(0)^3+f(y))=y=f(f(y)) \implies f(0)=0$$$P(x,0)$ $$f(x^3)=x^2f(x)$$$P(\sqrt[3]{x},f(y))$ $$f(x+y)=f(x)+f(y) \implies f \; \text{additive} \implies f(x)=x,-x \; \forall x \in \mathbb Z$$Call $Q(x,y)$ the assertion to the last replace. $Q((x+1)^3,(x-1)^3)$ $$2x^2f(x)+6f(x)=2f(x^3)+6f(x)=f(2x^3+6x)=f((x+1)^3+(x-1)^3)=(x+1)^2f(x+1)+(x-1)^2f(x-1)=2x^2f(x)+2f(x)+4f(1)$$Thus we have that $f(x)=xf(1)$ but since $f(1)=1,-1$ we have $f(x)=x,-x$ Assume that there exists $a,b \ne 0$ such that $f(a)=a$ and $f(b)=-b$. Then by $P(a,b)$ $$f(a^3-b)=a^3+b \implies a \; \text{or} \; b =0 \; \text{contradiction!!}$$Then $f(x)=x$ and $f(x)=-x$ are solutions and we are done

Jus the outline $x=0$ gives $f$ bijective. Use surjectivity to conclude that $f(0)=0$ Then we have $f(x^3)=x^2f(x)$ Then $y=f(y)$ gives $f$ is additive. Then with $x=x,y=-(x-1)^3$ and $x=(x+1)^3,y=-x^3$ And after equating for $3f(x^2)$ we conclude $f(x)=f(1)x$.And from the original f can be $x,-x$ Sorry for the initial wrong post.

@above f involution and additive doesnt imply f linear

primesarespecial wrote: Jus the outline $x=0$ gives $f$ bijective. Use surjectivity to conclude that $f(0)=0$ Then we have $f(x^3)=x^2f(x)$ Then $y=f(y)$ gives $f$ is additive. Then $f$ being additive and bijectives give $f$ linear which finishes. You can only conclude that $f$ is linear when the function is bounded,monotonic at any interval or continous at some point. $f$ being being involutive doesen't imply that $f$ is linear.

We prove that $f(x)=x$ and $f(x)=-x$ are the only solutions. Let $P(x,y)$ be the assertion in $f\left(x^3+f(y)\right)=x^2f(x)+y$. $P(0,x)$ : $$f(f(x))=x$$Hence we get involution and hence, $f$ is bijective. $P(1,f(0)) : f(0)=0$. $P(x,0) : f(x^3)=x^2f(x)$. Hence original equation reduces to $f(x^3+f(y))=f(x^3)+y$. $P(\sqrt[3]{x},f(y))$ in above equation : $$f(x+y)=f(x)+f(y)$$As, this holds $\forall x,y$, hence $f$ is additive. We have, $f(f(x))=x$. Replacing $x$ by $x^2$ gives: $f(f(x^2))=x^2\ge 0 $. Hence, $$f \text{ is bounded below on }[0,\infty)$$ Hence, $$ f\text{ additive} + f \text{ bounded} \implies f \text{ linear} $$Using additive, we get $f(x)=cx$ for some constant $c$. Using involution, we get $c^2=1 \implies c= 1,-1$. Hence, $f(x)=x$ and $f(x)=-x$. It can be easily checked that both $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$, satisfy given equations, and hence are our two solutions! Hence, we are done

$P(0,y) : f(f(y)) = y$. Let assume $f(a) = f(b)$. $P(x,a) , P(x,b) : a = b$ \implies $f$ is injective. $P(1,0) : f(1+f(0)) = f(1) \implies 1 + f(0) = 1 \implies f(0) = 0$. $P(x,0) : f(x^3) = x^2f(x) \implies f(f(x^3)) = f(x^2)f(f(x)) \implies x^3 = xf(x^2) \implies f(x^2) = x^2$. so for nonnegative $x$ we have $f(x) = \pm x$ which clearly both are answers. Now we will prove $f$ is odd. $P(x,f(y)) : f(x^3 + y) = x^2f(x) + f(y) \implies f(x^3+y) = f(x^3) + f(y)$. Now let $y = -x^3$ in last line we'll have $f(0)= f(x^3) + f(-x^3) \implies f(-x^3)= -f(x^3) \implies f$ is odd. Now for negative values of $x$, $f(x) = -f(-x)$. Answers : $f(x) = \pm x$.

Let $P(x,y)$ denote the given assertion. $P(0,x): f(f(x))=x$, so $f$ bijective. $P(1,0): f(1+f(0))=f(1)\implies 1+f(0)=1\implies f(0)=0$. $P(x,0): f(x^3)=x^2f(x)$. Thus, $f(x^3+f(y))=f(x^3)+y$, so $f$ is additive. Now we have $f((x+1)^3)=(x+1)^2f(x+1)$, so \begin{align*} f(x^3+3x^2+3x+1)=(x^2+2x+1)(f(x)+f(1)) \\ \implies f(x^3)+3f(x^2)+3f(x)+f(1)=x^2f(x)+2xf(x)+f(x)+x^2f(1)+2xf(1)+f(1) \\ \implies 3f(x^2)+3f(x)=2xf(x)+f(x)+x^2f(1)+2xf(1) \\ \end{align*} Now we also have $f((x-1)^3)=(x-1)^2f(x-1)$, so \begin{align*} f(x^3-3x^2+3x-1)=(x^2-2x+1)(f(x)-f(1)) \\ \implies f(x^3)-3f(x^2)+3f(x)-f(1)=x^2f(x)-x^2f(1)-2xf(x)+2xf(1)+f(x)-f(1) \\ \implies -3f(x^2)+3f(x)=-x^2f(1)-2xf(x)+2xf(1)+f(x) \\ \end{align*} We can add both equations now. We get \[6f(x)=2f(x)+4xf(1)\implies 3f(x)=f(x)+2xf(1)\implies f(x)=xf(1),\]so $f$ is linear. If we set $f(x)=ax$, then since $f$ involution, we have $a^2=1\implies a=1$ or $a=-1$. This gives the solutions $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$, which both work.

ftheftics wrote: See for any real $a$ we must get b so that ,$a=b^3$ we can say from here $f(a+x)=f(a)+f(x)$ rest proof is same as to Cauchy functional equation . That's why I don't write the rest . The induction in proving "Cauchy" won't work in reals (try it!), you need to do more. Hence your solution is (as @bora_olmez said) incomplete.

Mahdi_Mashayekhi wrote: $P(0,y) : f(f(y)) = y$. Let assume $f(a) = f(b)$. $P(x,a) , P(x,b) : a = b$ \implies $f$ is injective. $P(1,0) : f(1+f(0)) = f(1) \implies 1 + f(0) = 1 \implies f(0) = 0$. $P(x,0) : f(x^3) = x^2f(x) \implies f(f(x^3)) = f(x^2)f(f(x)) \implies x^3 = xf(x^2) \implies f(x^2) = x^2$. so for nonnegative $x$ we have $f(x) = \pm x$ which clearly both are answers. Now we will prove $f$ is odd. $P(x,f(y)) : f(x^3 + y) = x^2f(x) + f(y) \implies f(x^3+y) = f(x^3) + f(y)$. Now let $y = -x^3$ in last line we'll have $f(0)= f(x^3) + f(-x^3) \implies f(-x^3)= -f(x^3) \implies f$ is odd. Now for negative values of $x$, $f(x) = -f(-x)$. Answers : $f(x) = \pm x$. A lot of problems with this. You didn't do Pointwise Trap. You can't use injectivity like that. And involution implies bijectivity beforehand.

Let $P(x,y)$ denote the given assertion. $P(x,0)$ gives $f(f(x))=x,$ and in particular, bijectivity. Then $P(1,0)$ yields $f(0)=0.$ Compare $P(x,0)$ with $P(x,y)$ and with $P(x,f(y))$ to get $f$ is additive. Finally expand (with additivity) and compare $P(x+1,0)$ with $P(x-1,0)$ which readily gives $f(x)=xf(1).$ Checking shows $f\equiv \text{Id}$ and $f\equiv -\text{Id}$ clearly both work.

Call the assertion $P(x,y)$. $P(0,y)$ gives $f(f(y))=y$ (involution). So, $f$ is bijective. $P(1,0)$ gives $f(1+f(0))=f(1)\Longrightarrow 1=1+f(0)\Longrightarrow f(0)=0$. $P(x,0)$ gives $f(x^3)=x^2f(x)\Longrightarrow f(x^3+f(y))=f(x^3)+y$. We can replace $x^3$ by $x$, because $x^3$ also spans all real numbers. So, $f(x+f(y))=f(x)+y$. We can replace $f(x)$ with $x$ because $f(x)$ is surjective. We get $$f(x+y)=f(x)+f(y) \text{ --Cauchy's FE}.$$We now use the fact that $f(x^3)=x^2f(x)$. This gives $f((x+1)^3)=f(x^3)+3f(x^2)+3f(x)+f(1)=(x+1)^2(f(x)+f(1))=(x^2+2x+1)(f(x)+f(1))=x^2f(x)+2xf(x)+f(x)+x^2f(1)+2xf(1)+f(1)$, reducing to $$3f(x^2)=(2x-2)f(x)+(x^2+2x)f(1).$$$$\Longrightarrow3f((x+1)^2)=3f(x^2)+6f(x)+3f(1)=2xf(x+1)+(x^2+4x+3)f(1)=2xf(x)+6xf(1)+x^2f(1)+3f(1)$$$$\Longrightarrow 3f(x^2)=2xf(x)+6xf(1)+x^2f(1)-6f(x)=2xf(x)+x^2f(1)+2xf(1)-2f(x)$$$$\Longrightarrow f(x)=f(1)x=cx,c=f(1).$$Note that resubstituting gives $cx^3+c^2y=cx^3+y\Longrightarrow c=\pm1=f(x)$. Since $f(1)$ can have one distinct value only, we don't need to check for pointwise trap. $$\therefore\boxed{f(x)=x\ \forall\ x\in\mathbb{R},f(x)=-x\ \forall\ x\in\mathbb{R}}$$are the only solutions. $\square$

Let $P(x,y):=f(x^3+f(y))=x^2f(x)+y$ Claim: $f$ is injective Proof: Let $f(a)=f(b)$ $P(x,a)$ yields $f(x^3+f(a))=x^2f(x)+a$ $P(x,b)$ yields $f(x^3+f(b))=x^2f(x)+b$ Therefore $x^2f(x)+a=x^2f(x)+b\Longrightarrow a=b$, thus $f$ is injective, as desired $\square$. $P(0,0)$ yields $f(f(0))=0$ $P(0,0)$ yields $f(f(0)^3)=f(0)^2f(f(0))+f(0)\Longrightarrow f(f(0)^3)=f(0)\overset{\text{injectivity}}{\Longrightarrow}f(0)^3=0\Longrightarrow f(0)=0$ $P(0,x)$ yields $f(f(x))=x$ Furthermore $P(x,0)$ yields $f(x^3)=x^2f(x):=T(x)$ Thus $P(x,y)$ can be rewritten as $f(x^3+f(y))=f(x^3)+y:=Q(x,y)$ Now taking $Q(x,f(y))$ we obtain $f(x^3+f(f(y)))=f(x^3)+f(y)\Longrightarrow f(x^3+y)=f(x^3)+f(y)$ , also notice that if we take $x\to\sqrt[3]{x}$ we obtain $f(x+y)=f(x)+f(y)$ thus $f$ is additive. So now, $T(x+1)$ we obtain $f((x+1)^3)=(x+1)^2f(x+1)\Longrightarrow f(x^3+3x^2+3x+1)=(x^2+2x+1)(f(x)+f(1))$ Furthermore after expansion we obtain $f(x^3)+3f(x^2)+3f(x)+f(1)=x^2f(x)+x^2f(1)+2xf(x)+2xf(1)+f(x)+f(1)$ Which simplifies down to $3f(x^2)+2f(x)=x^2f(1)+2xf(x)+2xf(1)$ Moreover if we take $T(x-1)$ we obtain $f((x-1)^3)=(x-1)^2f(x-1)\Longrightarrow f(x^3-3x^2+3x-1)=(x^2-2x+1)(f(x)-f(1))$ Furthermore after expansion we obtain $f(x^3)-3f(x^2)+3f(x)=x^2f(x)-x^2f(1)-2xf(x)+2xf(1)+f(x)-f(1)$ Which simplifies down to $2f(x)-3f(x^2)=2xf(1)-2xf(x)-x^2f(1)$ Now notice that if we add $T(x+1)$ and $T(x-1)$ we obtain $4f(x)=4xf(1)\Longrightarrow f(x)=xf(1)\Longleftrightarrow f(x)=cx$, thus when we plug this result into our original functional equation we $f(x^3+cy)=cx^3+y\Longrightarrow cx^3+c^2y=cx^3=y\Longrightarrow c^2y=y$ thus $c=\pm 1$ or $f(x)=x\text{ and }f(x)=-x$ So now to avoid a pointwise trap let $f(a)=a\text{ and }f(b)=-b$ such that $a,b\neq0$ and by $P(a,b)$ we obtain $f(a^3-b)=a^3+b$ which forces either $a=0\text{ or }b=0$ which is a contradiction. In conclusion $\boxed{f(x)=x\text{ and }f(x)=-x, \forall x\in\mathbb{R}}$ $\blacksquare$.

Let $P(x, y)$ denote this assertion Claim: $f$ is an involution, and $f(0) = 0$ Proof Taking $P(0, y)$, we see that $f(f(y)) = y$ Now, $f(f(0)) = 0$, so taking $x = f(0)$, let $c = f(0)^3$, $f(f(y)+c) = y = f(f(y))$, and by injectivity (Involutions are injective), $c = 0$, so $f(0) = 0$ Claim: $f$ satisfies Cauchy's FE Proof $P(x, 0)$ implies $f(x^3) = x^2f(x)$, so replacing $x^2f(x)$ with $f(x^3)$ in the original FE, $f(x^3+y) = f(x^3) + f(y)$, but $x^3$ takes the values of all reals, thus $f$ satisfies Cauchy. Now, for rationals $x$, $f(x) = x$ or $f(x) = -x$. If $f$ works, so does $-f$, so assume WLOG $f(x) = x$ for all rationals. Then, consider some number $a$ such that $f(a) = b$. Then, $f(b) = a$. Since $f(x^3) = x^2f(x)$, $f(a^3) = a^2b$ and $f(b^3) = b^2 a$. So now, $f((a+b)^3) = (a+b)^2f(a+b) = (a+b)^3$. Now, we expand, so $a^2b+b^2a + 3a^3+3b^3 = a^3+b^3+3a^2b+3b^2a$, so $a = b$ or $a = -b$. But if $a = -b$, then $f(a) = -a$ and $f(a+q) = -a+q$, but then $f(a+q)$ should either be $-a-q$ or $a+q$, so this is a contradiction Thus, $f(x) = x$ and $f(x) = -x$ are the only solutions.