Let be: $$ P(x,y)\Rightarrow f (x ^ 3 + y ^ 3) = f (x ^ 3) + 3x ^ 3f (x) f (y) + 3f (x) (f (y)) ^ 2 + y ^ 6f (y) $$$$ P(0,0)\Rightarrow f(0)=f(0)+3f(0)^3\Rightarrow f(0)=0 $$$$ P(0,y)\Rightarrow f(y^3)=y^6f(y) $$Then: $$ Q(x,y)\Rightarrow f(x^3+y^3)=x^6f(x)+3x^3f(x)f(y)+3f(x)f(y)^2+y^6f(y) $$$$ Q(x,-x)\Rightarrow x^6f(x)+3x^3f(x)f(-x)+3f(x)f(-x)^2+x^6f(-x)=0...(1) $$$$ Q(-x,x)\Rightarrow x^6f(-x)-3x^3f(-x)f(x)+3f(-x)f(x)^2+x^6f(x)=0...(2) $$Sum $(1)$ and $(2)$ we get: $(f(x)+f(-x))(2x^6+3f(x)f(-x))=0$ if for some $x\in \mathbb{R}$ we have $3f(x)f(-x)=-2x^6$ then $x^6f(-x)-3f(-x)f(x)(x-3f(x))+x^6f(x)=0\Rightarrow x^6f(-x)+2x^6(x-3f(x))+x^6f(x)=0\Rightarrow f(-x)+2(x-3f(x))+f(x)=0$ $x^6f(x)+3f(x)f(-x)(x^3+f(-x))+x^6f(-x)=0\Rightarrow x^6f(x)+2x^6(x^3+f(-x))+x^6f(-x)=0\Rightarrow f(x)+2(x^3+f(-x))+f(-x)=0$ substract and we get: $3f(x)+f(-x)=-x^3+x$ and $3f(x)f(-x)=-2x^6\Rightarrow -2x^6+f(-x)^2=f(-x)(-x^3+x)$ and is easy to verify that this is impossible therefore: $$ f(x)=-f(-x) \ \forall \ x \in \mathbb{R} $$therefore: $$3f(x)^2(f(x)-x^3)=0 \ \forall x \in \mathbb{R} $$if $f(x)=0$ for some $x\in \mathbb{R}$ with $x\neq0$ then $Q(x,y)\Rightarrow f(x^3+y^3)=y^6f(y)$ if for $y\neq x$ we have $f(y)=y^3$ then $f(x^3+y^3)=y^9$ but $f(x^3+y^3)=0 \ \text{or} \ (x^3+y^3)^3$ then $y=0$ or $x=0$ contradiction, therefore $f(x)=0$ for all $x\in \mathbb{R}$ if $f(x)=x^3$ for some $x\neq 0$ suppose that exist $y\neq0$ such that $f(y)=0$ then $Q(x,y)\Rightarrow f(x^3+y^3)=x^6f(x)=x^9$ and It continues as in the previous case. Therefore, the solutions are: $\boxed{f(x)=x^3 \ \forall \ x\in\mathbb{R}} $ and $\boxed{f(x)=0\ \forall \ x\in\mathbb{R}}$

Let $P(x, y)$ be the assertion $f(x^3+y^3)=f(x^3)+3x^3f(x)f(y)+3f(x)(f(y))^2+y^6f(y)$ $P(0, 0):f(0)=f(0)+f(0)^3\implies f(0)=0$ $P(0, y):f(y^3)=y^6f(y)$ $P(x, y)-P(y, x):f(x^3)+3x^3f(x)f(y)+3f(x)(f(y))^2+y^6f(y) = f(y^3)+3y^3f(y)f(x)+3f(y)(f(x))^2+x^6f(x)$ $\implies 3x^3f(x)f(y)+3f(x)(f(y))^2 = 3y^3f(y)f(x)+3f(y)(f(x))^2 \implies \forall x, y: f(x)f(y)[x^3+f(y)-f(x)-y^3] = 0$ Let $f(a)\ne 0$ and $f(b)\ne 0$ then $f(a)-a^3=f(b)-b^3$ $\implies \exists c: \forall a:f(a)\ne 0:f(a)=a^3+c$ $f(a)\ne 0\implies a\ne 0\implies f(a^3)=a^6f(a)\ne 0$ $\implies a^9+c=f(a^3)=a^6f(a)=a^6(a^3+c)\implies c(a^6-1)=0$ If $c\ne 0$, then $\forall a$ not in $\{1, -1, 0\}:f(a)=0$ $P(\frac{1}{\sqrt[3]{2}}, \frac{1}{\sqrt[3]{2}}):f(1)=0$ $P(-\frac{1}{\sqrt[3]{2}}, -\frac{1}{\sqrt[3]{2}}):f(-1)=0$ $P(0, 0):f(0)=0$ $\boxed {\forall x: f(x)=0}$ 1) $\exists d:f(d) = 0$ and $d\ne 0$. $\forall y:f(y^3)=y^6f(y) \implies f(\sqrt[3]{d})=f(d^3)=0$ $P(\sqrt[3]{x}, \sqrt[3]{d}): f(x+d)=f(x)$ $P(x, -x): f(x^3+(-x)^3)=f(x^3)+3x^3f(x)f(-x)+3f(x)(f(-x))^2+(-x)^6f(-x)$ $\implies 0 = f(0)=x^6f(x)+3x^3f(x)f(-x)+3f(x)(f(-x))^2+x^6f(-x)$ $\implies \forall x \in \mathbb{R}, k\in \mathbb{Z} :0 = (x+kd)^6f(x+kd)+3(x+kd)^3f(x+kd)f(-x-kd)+3f(x+kd)(f(-x-kd))^2+(x+kd)^6f(-x-kd)$ $\implies 0 = (x+kd)^6f(x)+3(x+kd)^3f(x)f(-x)+3f(x)(f(-x))^2+(x+kd)^6f(-x)$ - this is polinom by k $\implies$ coefficient at $k^6$ equal to $0 \implies d^6[f(x)+f(-x)]= 0 \implies f(x)+f(-x)= 0$ $\implies$ coefficient at $k^3$ equal to $0 \implies 3d^3f(x)f(-x)= 0 \implies f(x)f(-x)= 0$ $ \implies 0= (f(x)+f(-x))^2-2f(x)f(-x) = f(x)^2+f(-x)^2\implies \boxed{\forall x:f(x) = 0}$ 2)$\forall d\ne 0:f(d)\ne 0 \implies \boxed{f(x)=x^3}$

I think in post 2 there is some typos since f(x) =x is not a solution

R-sk wrote: I think in post 2 there is some typos since f(x) =x is not a solution Oops, yep it's a typo. Thank you

The only solutions are $f(x)=x^3$ and $f(x)=0$. Let $P(x,y)$ be the given assertion. From $P(0,0)$ we have that $f(0)=0$. From $P(0,y)$ we have that $f(y^3)=y^6f(y)$. From $P(x,-x)$ we have that $x^6\left( f(x)+f(-x)\right)+3f(x)f(-x)\left( x^3+f(-x) \right)=0$. Now let $x \rightarrow -x$, then we have that $f(x)f(-x)\left(f(x) - x^3\right)=f(x)f(-x)\left( x^3+f(-x) \right)$ Let $S$ be the set of real numbers $x$ such that $f(x)=0$, in notation $S= \left\{ x \mid f(x)=0 \right\}$. Now assume that $card(S) \geq 2$. Then if $x\in S$, from $f(x^3)=x^6f(x)$, we have that $f\left( x^{3^{n}}\right) = 0$, for every integer $n$. Now from $P(x,x)$, where $x \in S$, we have that $f(2x^3)=0$ and again from $f(x^3)=x^6f(x)$, we have that $f\left( (x\sqrt[3]{2})^{3^n} \right) = 0$ Continuing on like that will leads us that for every positive real number $k$ we must have that $f(kx)=0$, when $x \in S$, thus we have that $f(t)=0$, for every number which has the same sign as every element in $S$. Now let's say that every element in $S$ was positive then, we must have that $f(x)=0$, for every positive $x$. Then we have that our assertion, when $x \geq 0$, is the following $f(x^3+y^3)=y^6f(y)=f(y^3)$, now choose $y$ to be negative and choose $x$ such that $x^3+y^3>0$, then we must have that $f(y^3)=0$, for any negative $y$. Thus for all real numbers we must have that $f(x)=0$. Now let's say that all were negative, then we must have that $f(x)=0$, for every negative $x$. Then we have that our assertion is turned into $f(x^3+y^3)=f(y^3)$, when $x$ is negative. Now choose a positive $y$ and a negative $x$ such that $x^3+y^3 > 0$. This easily implies that $f(y^3)=0$, for every positive $y$. Thus for all real numbers we must have that $f(x)=0$. Now assume that $card(S) = 1$, i.e. $f$ is injective at point $x=0$. From $f(x)f(-x)\left(f(x) - x^3\right)=f(x)f(-x)\left( x^3+f(-x) \right)$, we have that $f(-x)=f(x)-2x^3$. Now by plugging in $P(x,-y)$ in $P(y,-x)$ using $f(-x)=f(x)-2x^3$, we easily (we solve a quadratic and use trap point) get that $f(x)=x^3$, for all real $x$.

parmenides51 wrote: Find all functions $f :R \to R$ such that for all real numbers $x$ and $y$ $$f(x^3+y^3)=f(x^3)+3x^3f(x)f(y)+3f(x)(f(y))^2+y^6f(y)$$ My solution seems faster than the previous ones $\color{blue}\boxed{\textbf{Answer: }f\equiv x^3 \textbf{ or }0}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $$f(x^3+y^3)=f(x^3)+3x^3f(x)f(y)+3f(x)(f(y))^2+y^6f(y)...(\alpha)$$In $(\alpha) x\to 0, y\to 0:$ $$\Rightarrow f(0)=f(0)+3f(0)^3$$$$\Rightarrow f(0)=0$$In $(\alpha) x\to 0:$ $$\Rightarrow f(y^3)=y^6f(y)...(\beta)$$Replacing $(\beta)$ in $(\alpha):$ $$\Rightarrow f(x^3+y^3)=x^6f(x)+3x^3f(x)f(y)+3f(x)f(y)^2+y^6f(y)...(\theta)$$In $(\theta) y\to x:$ $$\Rightarrow f(x^3+y^3)=y^6f(y)+3y^3f(x)f(y)+3f(x)f(y)+x^6f(x)$$In $(\theta):$ $$\Rightarrow x^3f(x)f(y)+f(x)f(y)^2=y^3f(x)f(y)+f(x)^2f(y)...(\omega)$$$\color{red}\boxed{\text{If }\exists a\neq 0, b\neq 0 / f(a)\neq 0, f(b)\neq 0:}$ $\color{red}\rule{24cm}{0.3pt}$ In $(\omega) y\to a:$ $$\Rightarrow x^3f(x)f(a)+f(x)f(a)^2=a^3f(x)f(a)+f(x)^2f(a)...(\lambda)$$$$\Rightarrow x^3f(x)+f(x)f(a)=a^3f(x)+f(x)^2...(\lambda)$$In $(\lambda) x\to b:$ $$\Rightarrow b^3+f(a)=a^3+f(b)$$$$\Rightarrow f(b)=b^3+c, c \text{ is constant}$$Replacing in $(\alpha):$ $$\Rightarrow c=0$$$$\Rightarrow f(x)=x^3 \text{ or }0$$If $\exists c/f(c)=0$ In $(\alpha) x\to a, y\to c$ $$\Rightarrow f(a^3+c^3)=f(a^3)=a^9$$$$\Rightarrow (a^3+c^3)^3=a^9$$$$\Rightarrow c=0(\Rightarrow \Leftarrow)$$$$\Rightarrow f\equiv x^3$$$\color{red}\rule{24cm}{0.3pt}$ $\color{red}\boxed{\text{If }\not\exists a\neq 0 / f(a)\neq 0:}$ $\color{red}\rule{24cm}{0.3pt}$ $$\Rightarrow f\equiv 0$$$\color{red}\rule{24cm}{0.3pt}$ $$\Rightarrow \boxed{f\equiv 0 \text{ or }x^3}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$