I'm trying to show that for a torsion-free connection $\nabla$,
$R^{a}_{bcd}=\frac{2}{3}(R^{a}_{(bc)d}-R^{a}_{(bd)c})$
where $R^{a}_{bcd}$ is the Riemann curvature tensor (in abstract index notation i.e. no fixed coordinates)
$R^{a}_{bcd}Z^{b}X^{c}Y^{d}=(R(X,Y)Z)^{a}$
and the brackets in the subscript denote symmetrisation. Here's my attempt:
$T^{c}\nabla_{c}(T^{b}\nabla_{b}S^{a})=R^{a}_{bcd}T^{b}T^{c}S^{d} \quad $(Geodesic deviation, $\nabla_{T}\nabla_{T}S=R(T,S)T$)
$\implies R^{a}_{cbd}S^{d}=\nabla_{b}\nabla_{c}S^{a}$
$\implies R^{a}_{cbd}S^{d}-R^{a}_{bcd}S^{d}=\nabla_{b}\nabla_{c}S^{a}-\nabla_{c}\nabla_{b}S^{a}=R^{a}_{dbc}S^{d} \quad $ (Ricci identity)
$\implies R^{a}_{cbd}-R^{a}_{bcd}=R^{a}_{dbc}$
I've then tried to use symmetries of the Riemann tensor to get to the result, but to no avail. Any ideas? Thanks!