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

Answer

I'm posting an answer just to inform that the question has received an answer by Igor Rivin on MO.

https://mathoverflow.net/questions/143334/proving-sum-k-02m-1k-binom2mk3-1m-binom2mm-binom3mm

Mark Wildon mentioned that this is Dixon's identity.

http://en.wikipedia.org/wiki/Dixon%27s_identity