= Solution
Use the curvature convention $[\nabla_b,\nabla_c]V^d=R^d{}_{abc}V^a$, consistent with the coordinate formula later in the paper. Choose <normal coordinates> at an arbitrary point $p$. The <Levi-Civita connection> has $\Gamma^d{}_{ab}(p)=0$ and is torsion free, so $\Gamma^d{}_{ab}=\Gamma^d{}_{ba}$. At $p$,
$$
R^d{}_{abc}=\partial_b\Gamma^d{}_{ac}-\partial_c\Gamma^d{}_{ab}.
$$
Hence the cyclic sum is
$$
\begin{aligned}
R^d{}_{abc}+R^d{}_{bca}+R^d{}_{cab}
={}&\partial_b\Gamma^d{}_{ac}-\partial_c\Gamma^d{}_{ab}
+\partial_c\Gamma^d{}_{ba}-\partial_a\Gamma^d{}_{bc}\\
&+\partial_a\Gamma^d{}_{cb}-\partial_b\Gamma^d{}_{ca}=0.
\end{aligned}
$$
Each derivative cancels using symmetry of the two lower connection indices. Because the <Riemann curvature tensor> is antisymmetric in its last two indices, full antisymmetrization over $a,b,c$ equals one third of this cyclic sum. Thus
$$
\boxed{R^d{}_{[abc]}=0.}
$$
The statement is tensorial and $p$ was arbitrary, so it holds in all coordinates everywhere. This is the <first Bianchi identity>.
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