For a second-order rank, we know that the isotropic and deviatoric parts is given by $$ \frac{1}{2} \left( A_{ij} + A_{ji} \right) - \frac{1}{3} A_{ss} \delta_{ij} \, , $$ whereas for a third-rank tensor, it is given by $$ \frac{1}{6} \left( A_{ijk} + A_{ikj} + A_{jik} + A_{jki} + A_{kij} + A_{kji} \right) - \frac{1}{15} \left[ \left( A_{ssi} + A_{sis} + A_{iss} \right) \delta_{kj} + \left( A_{ssj} + A_{sjs} + A_{jss} \right) \delta_{ik} + \left( A_{ssk} + A_{sks} + A_{kss} \right) \delta_{ij} \right] \, . $$
Then, how can this be generalized for fourth- and fifth-rank tensors?
Thank you!
$ \def\tss{\mathop{\bullet}\limits} \def\dss{\mathop{\odot}\limits} \def\bbR#1{{\mathbb R}^{#1}} \def\a{\alpha}\def\b{\beta}\def\d{\delta}\def\t{\theta} \def\A{{\cal A}}\def\E{{\cal I}}\def\B{{\cal B}} \def\L{\left}\def\R{\right} \def\LR#1{\L(#1\R)} \def\BR#1{\Big(#1\Big)} \def\iso#1{\operatorname{iso}\LR{#1}} \def\dev#1{\operatorname{dev}\LR{#1}} \def\Diag#1{\operatorname{Diag}\LR{#1}} \def\sym#1{\operatorname{sym}\LR{#1}} \def\skew#1{\operatorname{skew}\LR{#1}} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\qiq{\quad\implies\quad} \def\c#1{\color{red}{#1}} $Let $\A$ denote an arbitrary $k^{th}$ order tensor, $\E$ the totally symmetric isotropic tensor of the same order, and $(\tss)$ the $k^{th}$ order contraction product, i.e. $$\A\tss\B = \sum_{i=1}^n\sum_{j=1}^n\ldots\sum_{\ell=1}^n \;{\A_{ij\ldots\ell}\;\B_{ij\ldots\ell}}$$
Then the isotropic part of $\A$ is defined as $$\eqalign{ \iso{\A} = \LR{\frac{\A\tss\E}{\E\tss\E}} \E \\ }$$ and the deviatoric part of the tensor is what's left over $$\eqalign{ \dev{\A} &= \A - \iso{\A}\qquad \\ }$$ Note that $\LR{\A\tss\E}$ and $\LR{\E\tss\E}\,$ are fully contracted scalar quantities.
Update
It occurs to me that you might be studying irreducible tensor decompositions, in which case you should probably study this post.