Hint:f injectif in 0 or null

If f injectif it easy to claim that f(x)=x

Answer: f(x)=0 or f(x)=x Let P(x,y) be the given equation. P(0,y) gives us f(0)=0. P(x,0) gives us f(x^3)=x*(f(x))^2 , so xf(x)>=0 for all real x. Case 1: If there exists a !=0, f(a)=0 P(x,a) tells us f(x^3)=xf(x+a)f(x-a)=x(f(x))^2 , so f(x+a)f(x-a)=(f(x))^2 for all real x.(1) For |x|<=|a|, we have (x+a)f(x+a)*(x-a)f(x-a)=(x^2-a^2)(f(x))^2. LHS>=0,RHS<=0, so LHS=RHS=0. For |x|<=2|a|,concern about (1),we can see f(x)=0....... Then it is easy to see f(x)=0 for all real x. Case 2: f(x)=0 is equivalent to x=0 For x !=0,P(x,x)gives us f(x)=x or -x. Combine with xf(x)>=0 , we can see f(x)=x for all real x.

The only solutions are $f\equiv 0$ and $f(x) = x$, which work. Now we show they are the only ones. Let $P(x,y)$ denote the given assertion. Assume that $f$ is not the identity. We will prove that $f$ is constant. $P(0,0): f(0) = 0$. $P(x,0): f(x^3) = xf(x)^2$. $P(x,x): f(x^3 - xf(x)^2) = 0$, so $f(x^3 - f(x^3)) = 0$. Since $x^3$ is surjective over reals, we have $f(x - f(x)) = 0$ for all reals $x$. Claim: If $f(x) \ne 0$, then $x$ and $f(x)$ have the same sign. Proof: Choose some $x$ with $f(x) \ne 0$. Clearly $x \ne 0$. Let $x = y^3$ for some real $y$ and we get that $x$ and $y$ are of the same signs. $P(y ,0): f(x) = y f(y)^2$. If $f(y) = 0$, then $f(x) = 0$, absurd. Thus, $f(y) \ne 0$, so $f(y)^2$ is positive, and therefore $f(x)$ has the same sign as $y$, which has the same sign as $x$. $\square$ Claim: If $f(x) = 0$, then $f(Nx) = 0$ for any positive integer $N$. Proof: It clearly suffices to show that for any $a,b$ with $f(a) = f(b) = 0$, we have $f(a+b) = 0$. $P(a+b, a): f((a+b)^3) = 0$. But this means $(a+b)f(a+b)^2 = 0$, so either $a + b= 0$ or $f(a+b) = 0$, both of which imply $f(a+b) = 0$. $\square$ Since $f$ isn't the identity, there exists $x$ with $f(x) \ne x$, but $f(x - f(x)) = 0$, so we have $c\ne 0$ with $f(c) = 0$. Claim: For any real number $r$, we have $f(r) = 0$. Proof: Suppose otherwise for some $r$. Clearly $r\ne 0$. For any positive integer $N$, $P(r, Nc): f(r^3) = r f(x+Nc) f(r - Nc)$. Hence $rf(r)^2 = rf(r + Nc)f(r- Nc) \implies f(r + Nc) f(r - Nc) = f(r)^2 > 0$. This also gives that $f(r + Nc)$ and $f(r - Nc)$ are nonzero. Now choose $N$ such that $r + Nc$ is positive and $r - Nc$ is negative (any $N$ with sufficiently large absolute value having the same sign as $c$ will satisfy this). By our second claim, we get that $f(r + Nc)$ is positive and $f(r - Nc)$ is negative, so $f(r + Nc) f(r - Nc) < 0$, absurd. $\square$

Hint : Consider the injectivity of $f$ in the point $0$.