About the symmetry of Riemann Tensor It is a problem in my homework.
First I was asked to show
$$
\nabla_a\nabla_bA_c-\nabla_b\nabla_aA_c=R_{a,b,c}^{\;\;\;\;\;d}A_d
$$
where $A$ is a (0,1)-tensor and $R_{a,b,c}^{\;\;\;\;\;d}$ is the Riemann curvature tensor, which is defined by
$$
\nabla_a\nabla_bV^c-\nabla_b\nabla_aV^c=R_{a,b,d}^{\;\;\;\;\;c}V^d
$$
I proved it.
Then the question says

Hence, show that $R_{a,b,c}^{\;\;\;\;\;d} +R_{b,c,a}^{\;\;\;\;\;d}+R_{c,a,b}^{\;\;\;\;\;d}=0$

However, I don't know how it can be deduced from the above identity. I tried to look at 
$$
[\nabla_a,\nabla_b]A_c+[\nabla_b,\nabla_c]A_a+[\nabla_c,\nabla_a]A_b
$$
But it seems not obvious that it is zero.
Can anyone help?

It suffices to show that
$$R_{a,b,c}^{d} A_{d} + R_{b,c,a}^{d}A_{d} + R_{c,a,b}^{d}A_{d} = 0$$
for any $A_d$. Expanding via the first formula, you get
$$\left( \nabla_{a}\nabla_{b}A_{c} - \nabla_{b}\nabla_{a}A_{c} \right) + \left( \nabla_{b}\nabla_{c}A_{a} - \nabla_{c}\nabla_{b}A_{a} \right)  + \left( \nabla_{c}\nabla_{a}A_{b} - \nabla_{a}\nabla_{c}A_{b} \right)$$
Now, if we take the first and last terms
$$\nabla_{a}\nabla_{b}A_c - \nabla_{a}\nabla_{c}A_b = \nabla_{a}(\nabla_{b}A_{c} - \nabla_{c}A_b) = \nabla_{a}(\nabla_{b}c(A) - \nabla_{c}b(A))$$
$$ = \nabla_{a}(\nabla_{b}c - \nabla_{c}b)A = \nabla_{a}([b,c]A) = 0$$
Since $a,b,c$ are coordinate vector fields for which the Lie bracket vanishes or is constant. The other pairs of terms cancel similarly.
