Simplifying tensor equation I was given a following tensor equation to simplify and solve:
$$A^s \cdot (A^s \otimes A^a \otimes I) \cdot A^s$$ where:
$A^s$- symmetric part of the representation; $A^s=\frac{1}{2}(A-A^T)$
$A^a$- anti-symmetric part of the representation; $A^a=\frac{1}{2}(A+A^T)$
$I$ - Kronecker delta, here
$I =\begin{pmatrix}
  1 & 0 & 0 \\
  0 & 1 & 0 \\
  0 & 0 & 1
 \end{pmatrix}$
$\otimes$ - Kronecker product
I have already calculated the $A^s$ and $A^a$ from the base tensor $A$ (in the exercise the basis is orthonormal, so $A^T = A_{ji}$):
$A =
 \begin{pmatrix}
  0 & 0 & 2\sqrt{3} \\
  0 & 6 & 0 \\
  0 & 0 & 2
 \end{pmatrix}$
$A^s =
 \begin{pmatrix}
  0 & 0 & \sqrt{3} \\
  0 & 6 & 0 \\
  \sqrt{3} & 0 & 2
 \end{pmatrix}$$
A^a =
 \begin{pmatrix}
  0 & 0 & \sqrt{3} \\
  0 & 0 & 0 \\
  -\sqrt{3} & 0 & 0
 \end{pmatrix}$
I have no idea what "simplifying" in this case means. Maybe there is some kind of relationship between $A^s$ and $A^a$ which I am not aware of and that's why it seems so unclear. Can you help me with the answer of provide tips on how to crack this problem?

I went to ask about this and got a solution. As all of the tensors here are $T^2$, using tensor properties, the whole equation can be simplified to:
$$(A^s \cdot A^s) \cdot A^a \cdot (I \cdot A^s) = \text{||using associative property||} = (A^s \cdot A^s)(I \cdot A^s) \cdot A^a$$
Because, as I said above, the tensors are $T^2$, then the "tensor multiplication" will reduce the row of the answer to $T^{|2-2|} = T^0 = R$ - it will be a scalar (where the result is $A\cdot B = A_{ij}B_{ij}$ - using summing convention). At the end we get multiplication of 2 scalars and multiplication of the product by the $A^a$ tensor:
$$(3+36+3+4)(6+2)\begin{pmatrix}
  0 & 0 & \sqrt{3} \\
  0 & 0 & 0 \\
  -\sqrt{3} & 0 & 0
 \end{pmatrix}=368\begin{pmatrix}
  0 & 0 & \sqrt{3} \\
  0 & 0 & 0 \\
  -\sqrt{3} & 0 & 0
 \end{pmatrix}$$
