This is q. 7 of ch. 4 from Do Carmo's book on Riemannian Geometry . Prove that: $$ \nabla R(X,Y,Z,W,T) + \nabla R(X,Y,W,T,Z) + \nabla R(X,Y,T,Z,W)=0.$$
Let $\{e_i\}$ a geodesic frame on $p$ , it is enough to prove that the identity for :
$$ \nabla R(e_i,e_j,e_k,e_l,e_h) + \nabla R(e_i,e_j,e_l,e_h,e_k) + \nabla R(e_i,e_j,e_h,e_k,e_l)=0.$$
The author says that the left side expression is :
$$ R(e_l,e_h,\nabla_{e_k} e_i,e_j) + R(e_h,e_k,\nabla_{e_l} e_i,e_j) + R(e_k,e_l,\nabla_{e_h} e_i,e_j).$$
I'd like to know why this is indeed the case.(you can check that: $$\nabla R(e_i,e_j,e_k,e_l,e_h)=\langle \nabla_{e_h} \nabla_{e_l} \nabla_{e_k}e_i - \nabla_{e_h} \nabla_{e_k} \nabla_{e_l}e_i +\nabla_{e_h} \nabla_{[e_k,e_l]}e_i , e_j \rangle).$$