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.
0. Let's prove more general form of Dixon's identity: $$ \sum(-1)^k\binom{a+b}{b+k}\binom{b+c}{c+k}\binom{c+a}{a+k}=\binom{a+b+c}{a,b,c} $$ (the version from OP corresponds to $a=b=c=n$).
1. Trinomial coefficients satisfy — and are defined by — a simple recurrence. So we only need to show that LHS — let's denote it $F(a,b,c)$ — satisfies $$ F(a,b,c)=F(a-1,b,c)+F(a,b-1,c)+F(a,b,c-1). $$
2. Let's rewrite LHS in 'Dyson style': it's the constant term of $$ (-1)^{a+b+c}\frac{(x-y)^{a+b}(y-z)^{b+c}(z-x)^{c+a}}{x^{2a}y^{2b}z^{2c}}. $$ But now the recurrence follows immediately from the identity $$ \frac{y^2}{(x-y)(y-z)}+\frac{z^2}{(y-z)(z-x)}+\frac{x^2}{(z-x)(x-y)}=-1 $$ (which is due to Euler, I believe).
$^\dagger$‘Good’ as in ‘I.J.Good. Short Proof of a Conjecture by Dyson’
// 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.
This is not a (full) answer, but it won't fit to a comment (and I hope it can be finished; but I also hope that there is a much simpler solution). The LHS is the coefficient $C_{2n}$ at $1$ in $((1-z)(1-w)(1-(zw)^{-1}))^{2n}$. We have $$C_n=\frac{1}{(2\pi i)^2}\oint\oint((1-z)(1-w)(1-(zw)^{-1}))^{n}\frac{dz}{z}\frac{dw}{w}$$ (we integrate over the unit circles). Let us compute $F(t):=\sum_n C_n t^n$. We just sum a geometric series and get
$$F(t)=\frac{1}{(2\pi i)^2}\oint\oint\frac{1}{zw-t(1-z)(1-w)(zw-1)}dz\,dw.$$ Let's compute the integral wrt. $z$ using residues. In the denominator we have a quadratic polynomial in $z$ with one root in the unit disk, so the residue is $\sqrt{}$ of its discriminant, and we get (modulo calculation mistakes) $$F(t)=\frac{1}{2\pi i}\oint\frac{1}{\sqrt{(w+t(w^2-1))^2-4t^2w(1-w)^2}}dw.$$ Under the square root we have a quartic polynomial in $w$, hence $F(t)$ is a complete elliptic integral of the first kind. Now we need to bring it to the normal form, or compute its Taylor series directly; at least for now I lack the energy to do it.
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.
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.
If this doesn't work, perhaps try induction? $m = 0$ is straightforward to check. However the induction bit will certainly hardcore.
– user88595 Sep 18 '13 at 14:08