How to show $\nabla (\Delta A) -\Delta(\nabla A)=\nabla Rm *A+Rm*\nabla A$? $A$ is a tensor field, $\nabla A(X,...)=(\nabla_XA)(...)$. Besides, $R$ is curvature operator and
$$
Rm(X,Y,Z,W)= \langle R(X,Y)Z,W\rangle
$$
and
$$
\Delta A = g^{ij}(\nabla _{X_i}\nabla _{X_j}A - \nabla_{\nabla_{X_i}X_j}A)
$$
For any tensor field $A,B$, $A*B$ is linear combination of tensor fields, each formed by starting with the tensor field $A\otimes B$, using metric to switch the type of components or to contractions.  Then, how to show
$$
\nabla (\Delta A) -\Delta(\nabla A)=\nabla Rm *A+Rm*\nabla A
~~~?
 \tag{1}
$$
PS: I know how to show $R(\cdot,\cdot)A=A*Rm$, for example $A=A_{ij}dx^i\otimes dx^j$, I can get
$$
[R(\cdot,\cdot)A]_{abcd}=-g^{lk}R_{abck}A_{ld}-g^{lk}R_{abdk}A_{cl}
$$
therefore, there is
$$
R(\cdot,\cdot)A=A*Rm
$$
But I fail to show (1).
 A: Firstly, let me restate the definition of the notation $A*B$ as a linear combination of various compatible contractions of $A \otimes B$. This notation is very permissive and absorbing, e.g.:
$$
- A*B \equiv A*B \tag{1}
$$
and
$$
A*B + A*B \equiv A*B \tag{2}
$$
where by $\equiv$ here I mean "can be renamed as".
Secondly, for the sake of easier calculations, let me switch to the abstract index notation (see e.g. R.Wald, General relativity, for an accessible introduction).
Thus, the Ricci calculus is presented by the following identities.
$$
2 \nabla_{[a} \nabla_{b]} X^c = R_{a b}{}^{c}{}_{d} X^d \tag{3}
$$
and
$$
2 \nabla_{[a} \nabla_{b]} \omega_d = - R_{a b}{}^{c}{}_{d} \omega_c \tag{4}
$$
You may be wondering about the sign conventions for the curvature, but for our needs it does not really matter, as we are going to rewrite the above equations as
$$
\nabla_a \nabla_b X^c = \nabla_b \nabla_a X^c + (R*X)_{a b}{}^c \tag{5}
$$
and
$$
\nabla_a \nabla_b \omega_c = \nabla_b \nabla_a \omega_c + (R*\omega)_{a b c} \tag{6}
$$
Since any tensor $A$ is can be presented as a linear combination of simple terms, that is tensor products of some amount of vectors and covectors, by the virtue of the linearity of the covariant derivative and using the Leibniz rule, we can see that
$$
\nabla_a \nabla_b A = \nabla_b \nabla_a A + (R*A)_{a b} \tag{7}
$$
for any tensor $A$ (this is the Ricci identities in disguise).
Now, down to the calculation.
$$
(\nabla \Delta A)_a - (\Delta \nabla A)_a  = \nabla_a \Delta A - \Delta \nabla_a A = \\
\nabla_a (g^{b c} \nabla_b \nabla_c A) - g^{b c} \nabla_b \nabla_c \nabla_a A = \\
g^{b c} ( \nabla_a \nabla_b \nabla_c A - \nabla_b \nabla_c \nabla_a A ) \equiv \\
g^{b c} ( \nabla_b \nabla_a \nabla_c A + (R*\nabla A)_{b a c} - \nabla_b \nabla_c \nabla_a A ) \equiv \\
g^{b c} ( \nabla_b \nabla_a \nabla_c A + (R*\nabla A)_{b a c} - \nabla_b ( \nabla_a \nabla_c A + (R*A)_{a c}) ) \equiv \\
g^{b c} ( \nabla_b \nabla_a \nabla_c A + (R*\nabla A)_{b a c} - \nabla_b \nabla_a \nabla_c A + (\nabla R*A)_{b a c} + (R * \nabla A)_{b a c} ) \equiv \\
(\nabla R*A)_a + (R*\nabla A)_a
$$
Let me emphasize, that the $*$-notation should be understood as
$$
\nabla \Delta A - \Delta \nabla A \equiv \nabla R*A + R*\nabla A
$$
