= Solution
Write $A_{ij}=\partial_j u_i$ and use <incompressibility>, $A_{ii}=0$. The displayed vector can be expressed as
$$
D_i=A_{ij}A_{jk}u_k-\frac12u_i\operatorname{tr}(A^2).
$$
On taking its <divergence>, $\partial_iA_{ij}=0$. Commuting the remaining <partial derivatives> gives
$$
A_{ij}(\partial_iA_{jk})u_k
=u_kA_{ij}\partial_kA_{ji}
=\frac12u_k\partial_k\operatorname{tr}(A^2),
$$
which cancels the derivative of the second term of $D_i$. The derivatives falling on $u_k$ leave
$$
\boxed{\partial_iD_i=A_{ij}A_{jk}A_{ki}=\operatorname{tr}(A^3)}.
$$
In <homogeneous turbulence>, averaging commutes with differentiation and $\langle D_i\rangle$ is position independent, provided these <moments> exist. Consequently $\langle\operatorname{tr}(A^3)\rangle=0$.
Using the <strain-rate tensor> and <vorticity> decomposition supplied in the question now gives the <Betchov relation>:
$$
\boxed{\langle S_{ij}S_{jk}S_{ki}\rangle=-\frac34\langle\omega_i\omega_jS_{ij}\rangle}.
$$
This uses homogeneity and <incompressibility>; isotropy is unnecessary.
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