 Solution

Let $m_I:X^3\to X$ add the coordinates indexed by a nonempty subset $I$ of the three-element index set. The Theorem of the Cube says that for every line bundle $\mathcal{L}$ on an abelian variety,is trivial, up to the harmless constant line given by the fiber of $\mathcal{L}$ at the identity.

Pull this line bundle back along $(f,g,h):Y\to X^3$. Pullback commutes with tensor products and duals, and $m_I\circ (f,g,h)$ is the corresponding sum of morphisms. The resulting bundle is precisely ${\mathcal{M}}_{f,g,h}$, so it is trivial.

Take $Y=X$, $f={\mathrm{id}}_X$, and let $g,h$ be the constant maps with values $x,y$. All pullbacks along constant maps are trivial line bundles. The formula for ${\mathcal{M}}_{f,g,h}$ then becomesor equivalently