The definition of a triple antisymmetrisation of a 3-rank tensor is defined as:

$$T_{[\mu\nu\rho]} = \frac{1}{6}(T_{\mu\nu\rho}+T_{\nu\rho\mu}+T_{\rho\mu\nu}-T_{\nu\mu\rho}-T_{\mu\rho\nu} - T_{\rho\nu\mu})$$

On the other hand the triple symmetrisation is defined as:

$$T_{(\mu\nu\rho)} = \frac{1}{6}(T_{\mu\nu\rho}+T_{\nu\rho\mu}+T_{\rho\mu\nu}+T_{\nu\mu\rho}+T_{\mu\rho\nu} + T_{\rho\nu\mu})$$

In textbook of Freedman & van Proeyen on Supergravity in equation (7.122) the 1. Bianchi identity is given as:

$$R_{\mu\nu\rho}^{~~~~~~~a} + R_{\nu\rho\mu}^{~~~~~~a} + R_{\rho\mu\nu}^{~~~~~~~a} = 0\tag{1} $$

If abreviated the identity is given as:

$$R_{[\mu\nu\rho]}^{~~~~~~~~~a}=0 \tag{A}$$

Why ?

Actually the Bianchi identity is half of the triple antisymmetrisation of a 3-rank +1 tensor, but it could also be seen as half of the triple symmetrisation of a 3-rank +1 tensor, i.e.

$$R_{(\mu\nu\rho)}^{~~~~~~~~~a}=0\tag{S}$$

For (A) to be true I would take (1), swap 2 adjacent indices, and substract the result from the original (1).

On the other hand, for (S) to be true I would take (1), swap 2 adjacent indices, and add the result to the original (1).

So I do not see the difference. However, if the Bianchi identity is used within a complicated expression, I guess, using one or the other form could lead to a very different result. What is going on here ?