= Solution
First use a genuine simple <bivector> of the specified chirality, with spinor part $\phi^{AB}=\alpha^A\alpha^B$. It is a <exterior product>: choose independent primed spinors $\rho^{A'},\sigma^{A'}$ and take $u^{AA'}=\alpha^A\rho^{A'}$, $v^{BB'}=\alpha^B\sigma^{B'}$. Their antisymmetric product is proportional to $w^{AA'BB'}=\alpha^A\alpha^B\epsilon^{A'B'}$.
For any antisymmetric $w^{ab}$, the <Riemann curvature tensor> acts on both tensor indices. In the printed curvature convention its complete contraction is
$$
[\nabla_a,\nabla_b]w^{ab}=R_{ab c}{}^a w^{cb}+R_{ab c}{}^b w^{ac}=-R_{bc}w^{cb}+R_{ac}w^{ac}=0,
$$
because the <Ricci tensor> is symmetric. Substituting the one-chirality <bivector> and the <chiral spinor curvature operator> decomposition, the primed-operator term vanishes on contraction with the symmetric $\alpha^A\alpha^B$. The other term is $2\Delta_{AB}(\alpha^A\alpha^B)$. Hence $\Delta_{AB}(\alpha^A\alpha^B)=0$.
Repeat with $\alpha+\beta$ and subtract the identities for $\alpha$ and $\beta$. Linearity leaves twice the mixed symmetric product, giving the <contracted chiral curvature identity>
$$
\boxed{\Delta_{AB}\big(\alpha^{(A}\beta^{B)}\big)=0.}
$$
The polarization works for spinor fields as well as pointwise spinors: the <chiral spinor curvature operator> is algebraic by the preceding part.
The displayed source expression $\alpha_A\beta_B\epsilon_{A'B'}$ needs this clarification. For independent $\alpha,\beta$ it is not antisymmetric in the spacetime indices. Its antisymmetric part is $\alpha_{(A}\beta_{B)}\epsilon_{A'B'}$, which is generally not simple. The proof above uses a genuinely simple bivector first and then polarization, so it establishes the requested conclusion without relying on the incorrect simplicity assertion.
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