Apologies in advance for this spectacularly uninteresting question, but it has just come up in my work. (Okay, not in a truly important way, but I am trying to gauge the scope of a certain construction.) 


Let $K$ be a field and $A$ be a division [Albert algebra](http://en.wikipedia.org/wiki/Albert_algebra) over $K$, i.e., a certain kind of $27$-dimensional commutative Jordan algebra over $K$. 



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> Is it true that for all nonzero elements $x,y \in A$, one has $(xy) y^{-1} = x$?
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Note that this is true in any finite-dimensional division algebra in which each subalgebra generated by two elements is associative, in particular in any composition algebra. But so far as I know (and by the way I know nothing about Albert algebras!), this "2-associativity" property does not hold here.



A silly technical question on Albert algebras
No. In an associative algebra $(xy+yx)y^{-1} + y^{-1}(xy+yx)= 2x+ yxy^{-1} + y^{-1}xy \neq 4x$ unless $x$ and $y$ commute. Hence, it does not hold even in special Jordan algebras.


