Find type of a differential form on an almost complex manifold If $M$ is a nearly Kähler manifold (that is, an almost Hermitian manifold on which $\nabla_X(J)X=0$) we have the three-forms
$$ A(X,Y,Z)=\langle\nabla_X(J)Y,Z\rangle \quad\text{and}\quad B(X,Y,Z)=\langle\nabla_X(J)Y,JZ\rangle. $$
How can I prove that these forms are of type $(0,3)+(3,0)$?

Edit: This claim can be found on pp. 3-4 of the paper "Nearly Kähler geometry and Riemannian foliations" by P. A. Nagy.
 A: First extend $J$ complex linearly so that $A$ and $B$ are defined on $TM\otimes_{\mathbb{R}}\mathbb{C}$. Note that $TM\otimes_{\mathbb{R}}\mathbb{C}$ decomposes as the direct sum of the two eigenspaces of $J$, namely $T^{1, 0}M$ and $T^{0,1}M$, with eigenvalues $i$ and $-i$ respectively.
Also extending the metric and $\nabla$ complex linearly, $A$ becomes a real three-form on $TM\otimes_{\mathbb{R}}\mathbb{C}$. As such, it can be uniquely written as 
$$A = A^{3,0} + A^{2, 1} + A^{1, 2} + A^{0, 3}$$
where $A^{p, q}$ is a $(p, q)$-form. In fact, as $A$ is a real form we have $A^{2,1} = \overline{A^{1,2}}$ and $A^{3,0} = \overline{A^{0,3}}$.
Note that $A(X, Y, JZ) = B(X, Y, Z) = -B(X, Z, Y) = -A(X, Z, JY) = A(X, JY, Z)$.  As $J$ preserves its eigenspaces, we see that the identity is true at the $(p, q)$ level.
Now suppose $X, Y \in \Gamma(M, T^{0,1}M)$ and $Z \in \Gamma(M, T^{1,0}M)$. Then
\begin{align*}
A^{1,2}(X, Y, JZ) &= A^{1,2}(X, JY, Z)\\
iA^{1,2}(X, Y, Z) &= -iA^{1,2}(X, Y, Z)\\
A^{1,2}(X, Y, Z) &= 0
\end{align*}
and hence $A^{1,2} = 0$ by skew-symmetry. As $A^{2,1} = \overline{A^{1,2}} = 0$, $A$ is of type $(3, 0) + (0, 3)$.
As $B(X, Y, JZ) = B(X, JY, Z)$, the same calculation can be used to show that $B$ is also of type $(3, 0) + (0, 3)$.

The form $A$ satisfies the identity $A(X, Y, JZ) = A(X, JY, Z)$. In fact, it follows from skew-symmetry that 
$$A(JX, Y, Z) = A(X, JY, Z) = A(X, Y, JZ),$$ 
and likewise for $B$. More generally, if $C$ is a $p$-form such that 
$$C(JX_1, X_2, \dots, X_k) = \dots = C(X_1, \dots, JX_i, \dots, X_k) = \dots = C(X_1, X_2, \dots, JX_k),$$
then $C$ is of type $(p, 0) + (0, p)$.

Thanks to the OP for helping me simplify the above arguments and for pointing out the final general fact.
