Showing Jacobi identity for Poisson Bracket We were given the following problem: show that $[A,[B,C]] + [B,[C,A]] + [C,[A,B]] = 0$ where $[A,[B,C]]$ et cetera are Poisson brackets. As I understand it this is a poisson bracket (where $\mathcal{H}$ is the Hamiltonian): $$\sum_i \left(\frac{\partial f}{\partial q_i} \frac{\partial \mathcal H}{\partial p_i} - \frac{\partial f}{\partial p_i} \frac{\partial \mathcal H}{\partial q_i} \right)$$
Well and good, that means, for example, that $[B,C]$ would be: $$
\sum_i \left(\frac{\partial B}{\partial q_i} \frac{\partial C}{\partial p_i} - \frac{\partial B}{\partial p_i} \frac{\partial C}{\partial q_i} \right)
$$
and $[A,[B,C]]$ would be $$\sum_{i,j} \frac{\partial A}{\partial q_j}\frac{\partial}{\partial p_j}\left(\frac{\partial B}{\partial q_i} \frac{\partial C}{\partial p_i} - \frac{\partial B}{\partial p_i} \frac{\partial C}{\partial q_i} \right)-\frac{\partial A}{\partial p_j}\frac{\partial}{\partial q_j}\left(\frac{\partial B}{\partial q_i} \frac{\partial C}{\partial p_i} - \frac{\partial B}{\partial p_i} \frac{\partial C}{\partial q_i} \right)$$ 
Multiplying this all out and taking the derivatives 
$$\sum_{i,j} \frac{\partial A}{\partial q_j}\left(\frac{\partial^2C}{\partial p_j p_i}\frac{\partial B}{\partial q_i} + \frac{\partial^2B}{\partial p_j q_i}\frac{\partial C}{\partial p_i}  - \frac{\partial^2 B}{\partial p_i p_j} \frac{\partial C}{\partial q_i} + \frac{\partial B}{\partial p_i} \frac{\partial^2 C}{\partial q_i} \right)-\frac{\partial A}{\partial p_j}\left(\frac{\partial B}{\partial q_i} \frac{\partial^2 C}{\partial p_i q_j} + \frac{\partial^2 B}{\partial q_i q_j} \frac{\partial C}{\partial p_i} - \frac{\partial^2 B}{\partial p_i q_j} \frac{\partial C}{\partial q_i} + \frac{\partial B}{\partial p_i } \frac{\partial^2 C}{\partial q_i q_j}\right)$$
its hard to keep track of this. So I figure that there has to be an easier way to do this, at least notation-wise. I have seen some other notations but there's never any explanation and I might be able to follow them if I understood what I was seeing. 
So for non-mathematicians who are seeing this in a physics class, if there are any suggestions as to a better notational system that would be most appreciated. 
I'm not really even sure what tags to put on this... 
 A: The algebra is lengthy, and I don't know any way around it, but for clarity, let's use this notaion:
$$
u_x=\frac{\partial u}{\partial x} \text{and}\ u_{xy}=\frac{\partial u}{\partial x\partial y}
$$
Since all $p_i$ and $q_i$ are independent for each value of $i$, let's constarain our focus to the problem of one dimension (i.e. all sums run from 1 to 1). Also, for clarity, $q=x$ and $p=y$.
$$
[A,B]=A_xB_y-A_yB_x\\
[A,[B,C]]=A_x(B_xC_y-B_yC_x)_y-A_y(B_xC_y-B_yC_x)_x
$$
Therefore,
$$
[A,[B,C]]=A_x(B_{xy}C_{y}+B_xC_{yy}-B_{yy}C_{x}-B_yC_{xy})-A_y(B_{xx}C_{y}+B_xC_{xy}-B_{xy}C_{x}-B_yC_{xx})\\
[B,[C,A]]=B_x(C_{xy}A_{y}+C_xA_{yy}-C_{yy}A_{x}-C_yA_{xy})-B_y(C_{xx}A_{y}+C_xA_{xy}-C_{xy}A_{x}-C_yA_{xx})\\
[C,[A,B]]=C_x(A_{xy}B_{y}+A_xB_{yy}-A_{yy}B_{x}-A_yB_{xy})-C_y(A_{xx}B_{y}+A_xB_{xy}-A_{xy}B_{x}-A_yB_{xx})
$$
By the symmetry of the problem, it will be enough to show that all of the terms multiplying $A_x$ sum to zero. Here is the sum of the terms multiplying $A_x$ from the second and third equations
$$
-B_xC_{yy}+B_yC_{xy}+C_xB_{yy}-C_yB_{xy}\\
=-(B_xC_{yy}-B_yC_{xy}-C_xB_{yy}+C_yB_{xy})\\
=-(B_{xy}C_{y}+B_xC_{yy}-B_{yy}C_{x}-B_yC_{xy})
$$
This is the negative of the terms multiplying $A_x$ in the first equation, so the solution does not depend on $A_x$ since its coefficient is $0$. Checking that the coefficients of $A_y,B_x,B_y, C_x, \text{and} \ C_y$ do in fact sum to zero is easy. Therefore,
$$
[A,[B,C]]+[B,[C,A]]+[C,[A,B]]=0
$$
