Proving $\sum_{k=0}^{2m}(-1)^k{\binom{2m}{k}}^3=(-1)^m\binom{2m}{m}\binom{3m}{m}$ (Dixon's identity) I found the following formula in a book without any proof:

$$\sum_{k=0}^{2m}(-1)^k{\binom{2m}{k}}^3=(-1)^m\binom{2m}{m}\binom{3m}{m}.$$

I don't know how to prove this at all. Could you show me how to prove this? Or If you have any helpful information, please teach me. I need your help.

Update : I crossposted to MO.
// I thought it would be nice to have an actual solution on this page (especially if that is not one of solutions posted @ MO).

OK, so we want to compute the coefficient
$$
[z^{2m}w^{2m}]\left\{(1-z)^{2m}(1-w)^{2m}(z-w)^{2m}\right\}
$$
i.e. the residue
$$
\operatorname*{res}_{z,w}\frac{(1-z)^{2m}(1-w)^{2m}(z-w)^{2m}}{z^{2m}w^{2m}}\frac{dz}z\frac{dw}w;
$$
after the substitution $Z=z/(1-w)$, $W=w/(1-z)$ (using that $z-w=\frac{Z-W}{1-ZW}$ and $\frac{dz}z\frac{dw}w=\frac{dZ\,dW}{ZW(1-ZW)}$) we see that it's equal to
$$
\operatorname*{res}_{Z,W}\frac{(Z-W)^{2m}}{(1-ZW)^{2m}Z^{2m}W^{2m}}\frac{dZ\,dW}{ZW(1-ZW)}=
[(ZW)^{2m}]\left\{\frac{(Z-W)^{2m}}{(1-ZW)^{2m+1}}\right\}
$$
which is equal to $(-1)^m\binom{2m}{m}\binom{3m}m$.

// The substitution looks somewhat magical. I've learned it from Gessel–Stanton paper mentioned by Todd Trimble @ MO.
Try this: use Newton's binomial on both sides of $$(1-x)^{2n}(1+y)^{2n}(1+z)^{2n}=[(1-x)(1+y)(1+z)]^{2n}$$

then track down and equate the coefficients at terms of the form $(xyz)^k$ on both sides. This should yield the required sum.