Manipulating Summations I am trying to manipulate an expression involving summations in order to take advantage of a particular identity, but my result is off by some factors of -1. How can I convert: 
$$f=\sum_{\kappa\psi\rho}(-1)^{2p+\psi+3r}\left(\begin{matrix}p&a&q\\\psi&\alpha&-\kappa\end{matrix}\right)\left(\begin{matrix}q&b&r\\\kappa&\beta&-\rho\end{matrix}\right)\left(\begin{matrix}r&c&p\\\rho&\gamma&-\psi\end{matrix}\right)$$
into something akin to:
$$f=A\sum_{\kappa\psi\rho}(-1)^{p-\psi+q-\kappa+r-\rho}\left(\begin{matrix}p&a&q\\\psi&\alpha&-\kappa\end{matrix}\right)\left(\begin{matrix}q&b&r\\\kappa&\beta&-\rho\end{matrix}\right)\left(\begin{matrix}r&c&p\\\rho&\gamma&-\psi\end{matrix}\right)?$$
Where $A$ is a multiplicative factor or some expression that does not involve summations (the whole point of this exercise is to eliminate the summations, and there is an identity that simplifies the sum in the second equation to a compact closed form expression).
If I multiply by $$(-1)^{-p+2\psi+q-\kappa-2r-\rho}$$ then do I have to multiply the other side of the equation by $$\sum_{\kappa\psi\rho}(-1)^{-p+2\psi+q-\kappa-2r-\rho}\;\;?$$
Note: $\left(\begin{matrix}a&b&c\\\alpha&\beta&\gamma\end{matrix}\right)$ is a Wigner 3j symbol. As such, the following relationships between the arguments to the 3j symbol hold:


*

*$0 \leq a, 0 \leq b, 0 \leq c$

*Any/all permutations of the triangular inequalities $\left| a-c \right| \leq b \leq a+b$

*$\left|\alpha\right| \leq a, \left|\beta\right| \leq b, \left|\gamma\right| \leq c$

*$\alpha+\beta+\gamma=0$


Additionally, in my case all arguments to the 3j symbol are integers.
 A: I'll use different variables for your condition $4$ so as not to get them mixed up with the $\alpha$, $\beta$, $\gamma$ in your summation. So $\left(\begin{matrix}x&y&z\\ \lambda&\mu&\nu\end{matrix}\right)$ vanishes unless $\lambda+\mu+\nu=0$, and hence the parity of $\lambda+\nu$ is that of $\mu$.
Thus, in the exponent of $-1$ you can replace $\psi-\kappa$, $\kappa-\rho$ and $\rho-\psi$ by $\alpha$, $\beta$ and $\gamma$, respectively. If these three expressions were linearly independent, you could eliminate all occurrences of $\psi$, $\kappa$ and $\rho$ in this way, and then take the parity factors outside of the summation. They aren't, however, since their sum is zero (which leads to a constraint $\alpha+\beta+\gamma=0$; the sum vanishes unless this is fulfilled).
But you can still use these replacements to rewrite the sum. With $\psi=\psi+\psi-\psi=(\kappa-\alpha)+(\rho+\gamma)-\psi$, you get 
$$
\begin{eqnarray}
\sum_{\kappa\psi\rho}(-1)^{2p+\psi+3r}\,\Gamma
&=&
(-1)^{2p+3r}\sum_{\kappa\psi\rho}(-1)^{\psi}\,\Gamma\\
&=&
(-1)^{2p+3r}\sum_{\kappa\psi\rho}(-1)^{(\kappa-\alpha)+(\rho+\gamma)-\psi}\,\Gamma\\
&=&
(-1)^{2p+3r+\gamma-\alpha}\sum_{\kappa\psi\rho}(-1)^{\kappa+\rho-\psi}\,\Gamma\\
&=&
(-1)^{2p+3r+\gamma-\alpha}\sum_{\kappa\psi\rho}(-1)^{-\kappa-\rho-\psi}\,\Gamma
\end{eqnarray}
$$
Here $\Gamma$ is the product of the three $3j$ symbols. (I'm not sure why you have $2p$ and $3r$ in the exponent, since these are equivalent to $0$ and $r$, respectively, if $p$ and $r$ are integers, but I've left them as you had them.)
So the factor $A$ that you're looking for is $(-1)^{p+q+\alpha+\gamma}$.
