What was the genesis of Hua's identity? Many resources I have read prove Hua's identity more-or-less mechanically. I have seen there is more than one raison d'être for Hua's identity: e.g. its connection to the fundamental theorem of projective geometry, and also Jordan algebra theory. My impression, though, is that these two things are mostly application rather than inspiration. (I could be wrong, though.)

I would very much like to know how Hua's identity arose, hopefully with motivation/intuition as to how it was discovered.

I have intended to get ahold of the(?) original proof by Hua in hopes that it contained such information, but so far I haven't managed to lay my hands on the original citation(s). This would be a much-appreciated bonus to any solution.
If it turns out there is a good retroactive motivation/intuition for deriving the identity that beats the original, of course that would be welcome as well.

Happily I've seen the original paper now (thanks Martin). Surprisingly, the identity cited by all authors since the paper is different-looking from the original. I will have to compare the two versions and see if this version gives any more insight. No direct intuition about its origins are apparent, and indeed it is called "nearly trivial" although it seems a bit mystifying, IMO.
 A: Well, I can only guess, but if I were Hua this is what I would have thought to derive the identity and say it is "almost trivial":
1) We want to generalize the theorem of Cartan and Dieudonné (now called Cartan-Brauer-Hua), so we want to express an element $a$ as a combination of sums, products and inverses of conjugates of $a,b$, for any other $b$ (such that $ab\neq ba$).
2) Immediately we think about proving our luck with the well known formula for multiplicative commutators in division rings, since it can readily give us $a$ as a factor so that we can solve it "as a fraction", and it is close to conjugations: If we denote $(a,b):=a^{-1}b^{-1}ab$ then
$$a(a,b)=(a-1)(a-1,b)+1.$$
Therefore $$a((a,b)-(a-1,b))=1-(a-1,b).$$
3) If we expand we see we don't yet have conjugates due to the $b$ factors at the right of $(a,b)=a^{-1}b^{-1}ab$, etc.; but since it is the same for all terms, we add a right $b^{-1}$ to solve the problem, and get
$$a((a,b)-(a-1,b))b^{-1}=(1-(a-1,b))b^{-1}$$
$$a(a^{-1}b^{-1}a-(a-1)^{-1}b^{-1}(a-1))=b^{-1}-(a-1)^{-1}b^{-1}(a-1)$$
and now Hua's identity follows by solving for $a$.
