Let $a,b,c$ be basis vectors over some commutative associative algebra X. Consider the algebra A spanned by these basis vectors over X, with the multiplication rules $pq=q$ for all $p,q \in \{a,b,c\}$, and extend to all of A by requiring that the multiplication is bilinear over X. Thus $(ua+vb+wc)(xa+yb+zc)=(u+v+w)(xa+yb+zc)$

Write $S(xa+yb+zc)=x+y+z \in X$ for convenience, so that one can write $xy=S(x)y$ for all $x,y \in A$

This gives an associative multiplication, since $(xy)z=S(x)yz=x(yz)$. Now since $ab - ba = b-a$ the algebra isn't commutative. But if I write $xyz+yzx+zxy-yxz-zyx-xzy$ we have $xyz=S(x)S(y)z=yxz$. Thus $P_3(x,y,z)=0$ for all $x,y,z \in A$

Let $a,b,c$ be basis vectors over some commutative associative algebra $X$. Consider the algebra $A$ spanned by these basis vectors over $X$, with the multiplication rules $pq=q$ for all $p,q \in \{a,b,c\}$, and extend to all of $A$ by requiring that the multiplication is bilinear over $X$. Thus $(ua+vb+wc)(xa+yb+zc)=(u+v+w)(xa+yb+zc)$.

Write $S(xa+yb+zc)=x+y+z \in X$ for convenience, so that one can write $xy=S(x)y$ for all $x,y \in A$.

This gives an associative multiplication, since $(xy)z=S(x)yz=x(yz)$. Now since $ab - ba = b-a$ the algebra isn't commutative. But if I write $xyz+yzx+zxy-yxz-zyx-xzy$ we have $xyz=S(x)S(y)z=yxz$. Thus $P_3(x,y,z)=0$ for all $x,y,z \in A$.