Maybe is $ P^{3}(x)+3P^{2}(x)=P(x^3)+3P^{2}(-x) $...

No,it's not.But there was a mistaken sign.I corrected it.Thanks for the comment.

Does anyone have solutions??

If P is constant we obtain as a solutions $P=-1$ and $P=-2$. In what will follow we presume that $deg(P)\geq 1$ Let $x=0$ and $y=P(0)$; we obtain $y^3+3y^2=y-3y$ with sollutions $y=0, y=-1,y=-2$. Case 1: $y=0$ we obtain $P(x)=x^kQ(x)$ with $Q(0)\neq 0$. Introducing in the initial condition we obtain: $x^{3k}Q(x^3)+3x^{2k}Q(x)=x^{3k}Q(x^3)-3(-x)^{k}Q(-x)$; after we simplify with $x^k$ we obtain $x^{2k}Q(x^3)+3x^{k}Q(x)=x^{2k}Q(x^3)-3(-1)^{k}Q(-x)$, relation true for $x\neq 0$. But Q is a polynomail function so continuous. As a consquence that relation is true also for $x=0$. we obtain $Q(0)=0$, false. Case 2: in the same way we show that $P(0)\neq -2$ Case 3: $P(0)=-1$. Then exists Q such that $P(x)=x^kQ(x)-1$ and $Q(0)\neq -1$ We wil show that k is an even number and $Q=1$ If we introdu ce in the condition the new form of P we obtain $x^{3k}Q^3(x)-3x^{2k}Q^2(x)+3x^kQ(x)-1+3x^{2k}Q^2(x)-6x^kQ(x)+3=$ $x^{3k}Q^3(X)-1-3((-X)^kQ(-x)-1)$ after calculations and simplification with $x^k$ we obtain $x^{2k}Q^3(X)-3Q(X)=x^{2k}Q(x^3)+3(-1)^{k+1}Q(-x)$. If k is odd, taking $x=0$ we obtain $-3Q(0) = 3Q(0)$ so $Q(0)=0$ false. So $P(x)=x^{2k}Q(x)-1$ To show that Q is 1 we introduce the new form of P in the condition. After calculus we obtain: $x^{4k}(Q^3(x)-Q(x^3))= 3(Q(x)-Q(-x))$ If $Q(x) =\sum_0^{2n} a_i x^{i}$ then $Q(-x) =\sum_0^{2n} a_i (-x)^{i}$ and $Q(x)-Q(-x)=a_1x+a_3x^3+\cdot\cdot\cdot+a_{2p+1}x^{2p+1}$, where $2p+1$ is the highest number suchh that $a_{2p+1}\neq 0$ From rhe condition we obtain $a_1=a_3=\cdot\cdot\cdot = a_{4k-1}=0$, so $Q(x)-Q(-x)=a_{4k+1}x^{4k+1}+a_{4k+2}x^{4k+3}+\cdot\cdot\cdot+a_{2p+1}x^{2p+1}$. After simplifications we obtain $Q^3(x)-Q(x^3)=a_{4k+1}x^{4k+1}+\cdot\cdot\cdot+a_{2p+1}x^{2p+1}$ In the left side the coefficient of $x^{4n+29+1}$ is $3a_{2n}^2a_{2p-1}\neq 0$ and in the right one is $0$. In the same way we demonstrate in the case of $deg(Q) = 2n+1$ So Q must be a constant. As a solution for Q we obtain $Q=0$ so $P=-1$ $Q=-1$ unacceptable and $Q=1$ so $P(x)=x^{2k}-1$

I think you missed the solution $P(x)=-x^{2k}-1.$

You are wright; I was in a hurry this solution is obtained for $Q=-1$

kukyrami wrote: In the left side the coefficient of $x^{4n+29+1}$ is $3a_{2n}^2a_{2p-1}\neq 0$ and in the right one is $0$. In the same way we demonstrate in the case of $deg(Q) = 2n+1$ So Q must be a constant. As a solution for Q we obtain $Q=0$ so $P=-1$ $Q=-1$ unacceptable and $Q=1$ so $P(x)=x^{2k}-1$ Here we obtain $a_1=a_3=...=a_{2p+1}=0$=>$Q(x)=Q(-x)$ => $Q(x^3)=Q(x)^3$=>$Q(x)=-x^{2d}$ or $Q(x)=x^{2d}$=>$P(x)=-x^{2k}-1$ or $P(x)=x^{2k}-1$