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  "body": "\n## Statement\n\n1. For every nonnegative integer $a$, the [[binomial coefficients|0GBZM0MT0F]] satisfy\n$$\\sum_{k = -a}^{a} (-1)^k \\binom{2a}{a + k}^3 = \\frac{(3a)!}{(a!)^3}.$$\n2. For nonnegative integers $a, b, c$, with $k$ running over all integers and $\\binom{n}{j}$ read as $0$ for $j \u003c 0$ or $j \u003e n$,\n$$\\sum_{k} (-1)^k \\binom{a + b}{a + k} \\binom{b + c}{b + k} \\binom{c + a}{c + k} = \\frac{(a + b + c)!}{a!\\, b!\\, c!}.$$\n3. Let $a, b, c$ be complex numbers such that $1 + a - b$, $1 + a - c$ and $1 + \\frac{a}{2}$ are not in $\\set{0, -1, -2, \\dots}$ and $\\operatorname{Re}\\left( \\frac{a}{2} - b - c \\right) \u003e -1$.\nThen, with $\\Gamma$ the [[gamma function|0H5SYBSED9]] and ${}_3F_2$ the [[generalized hypergeometric series|VDBAW9PN88]],\n$${}_3F_2\\left( \\begin{matrix} a,\\ b,\\ c \\\\ 1 + a - b,\\ 1 + a - c \\end{matrix} ; 1 \\right) = \\frac{\\Gamma(1 + \\frac{a}{2})\\, \\Gamma(1 + a - b)\\, \\Gamma(1 + a - c)\\, \\Gamma(1 + \\frac{a}{2} - b - c)}{\\Gamma(1 + a)\\, \\Gamma(1 + \\frac{a}{2} - b)\\, \\Gamma(1 + \\frac{a}{2} - c)\\, \\Gamma(1 + a - b - c)}.$$\n\n## Remarks\n\nPart 1 is the case $a = b = c$ of part 2.\nThe ${}_3F_2$ in part 3 is *well-poised*: each upper parameter plus the matching lower parameter gives $1 + a$.\nPart 3 is the case $d = \\frac{a}{2}$ of the summation of a well-poised ${}_5F_4$.\nThat series has upper parameters $a, 1 + \\frac{a}{2}, b, c, d$ and lower parameters $\\frac{a}{2}, 1 + a - b, 1 + a - c, 1 + a - d$.\nFor $d = \\frac{a}{2}$, two upper and two lower parameters cancel.\nThe condition on $\\operatorname{Re}\\left( \\frac{a}{2} - b - c \\right)$ is exactly the condition for the series at $1$ to converge.\nFor $a = 1$ in part 1, the sum is $-1 + 8 - 1 = 6 = 3!/(1!)^3$.\n\n## Explanations\n\n### L1 human\n\nTake row $2a$ of Pascal's triangle, cube each entry, and add the cubes with alternating signs.\nDixon's identity says the result is the multinomial coefficient $\\frac{(3a)!}{(a!)^3}$, the number of ways to split $3a$ objects into three groups of $a$.\nFor $a = 1$ the row is $1, 2, 1$, and $-1 + 8 - 1 = 6$.\n\n### L2 human\n\nThe binomial forms are the terminating cases of the hypergeometric form in part 3.\nPart 3 is one of the classical closed-form evaluations of ${}_3F_2(1)$, next to the Pfaff–Saalschütz theorem for balanced series and Watson's and Whipple's sums.\nDixon's identity also has a short proof by the Wilf–Zeilberger method of certificate proofs.\n",
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