By a ring I mean a $\mathbb{Z}$-module $A$ equipped with a binary operation $*:A\times A\rightarrow A$ that is $\mathbb{Z}$-bilinear and denoted by $(x,y)\mapsto xy$.
Let $A$ be a ring.
$A$ is said to be right alternative if $(xy)y=x(yy)$ for any $x,y\in A$.
$A$ is said to be left alternative if $(xx)y=x(xy)$ for any $x,y\in A$.
$A$ is said to be flexible if $(xy)x=x(yx)$ for any $x,y\in A$.
I know that, if we assume two of the properties above, then the remaining property holds.
A ring is said to be alternative if all three properties above hold.
However, I do not know an easy example of a right alternative ring that is not alternative.
Of course I can try the free right alternative ring, but are there easier examples? There is another very complicated example, provided by Mikheev in the article Simple Right Alternative Rings, that is a simple right alternative ring that is not alternative.
I know that every right alternative ring is power-associative, where a ring $A$ is said to be power-associative if for every $x\in A$ the subring $A(x)$ generated by $x$ is associative. Therefore, in a power-associative ring, we can define powers of an element as usual.
I know that, by a result from Mikheev, a right alternative ring satisfies the identity $(x,x,y)^4=0$, where $(x,y,z)$ stands for $(xy)z-x(yz)$. In particular, every right alternative ring without nonzero nilpotent elements is alternative.
I know that, by a result from Albert, every finite-dimensional right alternative algebra over a field of characteristic $\neq2$, which does not contain nil-ideals, is alternative
I also know that, by another result from Albert, every semisimple, right alternative algebra over a field of characteristic $0$ is alternative.