We have the following beautiful result for Pauli $su(2)$ matrices
$$(\vec{\sigma}\cdot\vec{a})(\vec{\sigma}\cdot\vec{b}) = \mathbb{I} ~\vec{a}\cdot\vec{b} + i (\vec{a} \times \vec{b}) \cdot \vec{\sigma}.$$
Do we have a similar structure for Gell-Mann $su(3)$ matrices? Specifically, what would the following be
$$(\vec{\lambda}\cdot\vec{a})(\vec{\lambda}\cdot\vec{b}) = ~?$$
Yes, of course. The anticommutator for Gell-Mann matrices is somewhat more elaborate than for Pauli matrices, as there is also a d-coefficient, so splitting the $\lambda$-matrix bilinear into commutators and anticommutators yields $$ (\vec{\lambda}\cdot\vec{a})(\vec{\lambda}\cdot\vec{b}) = a^\mu \lambda^\mu ~b^\nu \lambda^\nu = a^\mu b^\nu \left (\tfrac{1}{2} [\lambda^\mu,\lambda^\nu] + \tfrac{1}{2} \{\lambda^\mu,\lambda^\nu \}\right )= \\ =a^\mu b^\nu ( if_{\mu \nu\kappa} \lambda^\kappa + d_{\mu\nu\kappa} \lambda^\kappa + \tfrac{2}{3} \delta_{\mu\nu} 1\!\!1) \\ =\tfrac{2}{3} 1\!\!1 a\cdot b +a^\mu b^\nu (if_{\mu \nu\kappa}+d_{\mu \nu\kappa})\lambda^\kappa, $$ the second term being analogous to the cross-product, except now it has both an antisymmetric and a symmetric piece.
Bonus point. Combining two octets will yield a reducible 64, $$8\otimes 8= 27\oplus\overline{10}\oplus10\oplus8\oplus8\oplus1 .$$ The symmetric singlet is explicit above (just as the SU(2) singlet for the Pauli matrices is), and the symmetric d term above reduces to one of the two 8 s, and not the 27.
The antisymmetric f term reduces to the other 8 and not the 10 and its conjugate.
The 8 hermitian matrices $m^\kappa _{\mu\nu}\equiv (if_{\mu\nu\kappa} +d_{\mu\nu\kappa} )$ are very sparse, much more so than their SU(2) angular momentum analogs. Their (imaginary) antisymmetric piece vanishes unless there are 1 or 3 indices from the set 2,5,7; and their (real) symmetric piece vanishes unless there is an even number of indices from the same set. For instance, $$ m^2= \begin{pmatrix} 0 & 0 & -i &0 &0 &0 &0 &0 \\ 0 & 0 & 0 &0 &0 &0 &0 &1/\sqrt{3} \\ i & 0 & 0 &0 &0 &0 &0 &0 \\ 0 & 0 & 0 &0 &0 &i/2 &-1/2 &0 \\ 0 & 0 & 0 &0 &0 &1/2 &i/2 &0 \\ 0 & 0 & 0 &-i/2 &1/2 &0 &0 &0 \\ 0 & 0 & 0 &-1/2 &-i/2 &0 &0 &0 \\ 0 & 1/\sqrt{3} & 0 &0 &0 &0 &0 &0 \end{pmatrix} , $$ and so on. Note this matrix is only 3/16 full!
(Unfortunately) there is no such generalization: the properties of the Pauli matrices that make such identities possible are closely tied to the $\mathbb{Z}_2\times \mathbb{Z_2}$ graded structure of the matrices (see 2. below).
However, as part of this negative answer I will point you to the following:
The first will give you geometric relations similar to the cross product of Pauli matrices, and also a $\star$ operation on the Gell-Mann matrices, but not what you want. The second will provide you with an alternate basis (of non-hermitian but unitary matrices) that nevertheless have some nice properties (such as $A^3\sim 1_{3\times 3}$, which generalize some of the properties of the Pauli’s.
(I wish someone can show my answer to be wrong as I’d love to know such a relation.)
