Does p^3(x) mean it is multiplied 3 times or it is a composite function

It is same as Quote: Find all polynomials $P(x)$ with real coefficents, such that for all $x,y,z$ satisfying $x+y+z=0$, the equation below is true: \[P(x)^3+P(y)^3+P(z)^3=3P(xyz)\] $z=0,y=-x: P(x)^3+P(-x)^3=3P(0)-P(0)^3=c \to P(-x)^3=c-P(x)^3$ $y=1-x,z=-1:P(x)^3+P(1-x)^3+P(-1)^3=3P(x(x-1))=P(x)^3+c-P(x-1)^3+P(-1)^3$ Let $deg(P(x))=n>0$ then $deg(P(x)^3-P(x-1)^3)= 3n-1$ and $deg(P(x(x-1))=2n \to n=1$ For $n=0:P=0,-1,1$ For $n=1:P=ax+b$ and so $(ax+b)^3+(ay+b)^3+(-a(x+y)+b)^3=3(-axy(x+y)+b)$ $x=y=0 \to 3b^3=3b \to b=0,-1,1$ Case 1: $b=0: a^3x^3+a^3y^3-a^3(x+y)^3=-3axy(x+y)$ $3a^3(xy(x+y))=3axy(x+y) \to a=\pm 1$ Case 2: $b= \pm 1: (ax \pm 1)^3+(ay \pm 1)^3+(-a(x+y) \pm 1)^3=3(-axy(x+y)\pm 1)$ $y=0,x=1: (a \pm1)^3 \pm1+(-a\pm1)^3=\pm 3 \to 6a^2 =0$ - contradiction So answer: $P(x)=0, \pm 1, \pm x$

$x, y, z \to 0$ : $P(0)=0, \pm 1$. $x \to (y+z)$ : $P(-y)^3 + P(-z)^3 + P(y+z)^3 = 3P(yz(y+z)$ for all $y,z$. $y \to -z$ : $(P(z)+P(-z))(P(z)^2 - P(z)P(-z)+P(-z)^2)=2P(0)$. Note that the degree of $P(z)^2 - P(z)P(-z)+P(-z)^2$ is equal to twice of $P(z)$. Therefore, $P(z)+P(-z)=0$ and $P(0)=0$. Let $P(x)=a_n x^{2n+1} +... + a_0 x$. Then, $(a_n)^3 ((y+z)^{6n+3} - y^{6n+3} - z^{6n+3})$ are the greatest degree terms on the left side and $(a_n)(yz(y+z))^{2n+1}$ are the greatest degree terms on the right side. If $n>0$, the term $y^{6n+2} z$ survives on the left but not on the right. Therefore, $n=0$ and ${a_n}^3 =a_n$. Thus, we get $P(x)=\pm x , 0$.