$(ab)c + a(bc) = 2 b (ac) \implies^? x(yz) = (xy)z$?

Question

Consider some unital commutative algebra $A$ such that for all its elements we have

$$(ab)c + a(bc) = 2 b (ac) $$

Does this imply the algebra is associative ?

or in symbols :

$$(ab)c + a(bc) = 2 b (ac) \implies^? x(yz) = (xy)z$$

Answer 1

The following shows that the algebra is associative whenever the multiplication is distributive and 6 is not a zero-divisor.

Taking $a=b$ gives $(aa)c=a(ac)$. Next taking $a=b=x+y$ and $c=z$ in this new relation and using commutativity and distributivity gives $$ (x^2+y^2+2xy)z = (x+y)(xz+yz) = x(xz) + y(yz) + x(yz)+ y(xz), $$ so that $$ x(yz)+y(xz) = 2(xy)z. $$ Finally, we have $$ 2b(ac) - a(bc) = (ab)c = (ba)c = 2a(bc) - b(ac), $$ so that $$ 3a(bc) = 3b(ac). $$ Thus, as 6 is a not a zero divisor, we get $x(yz)=y(xz)$, which combined with the earlier relation gives $2x(yz)=2(xy)z$, and hence $x(yz)=(xy)z$.