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Apr 29, 2022 at 14:12 history edited Denis Serre CC BY-SA 4.0
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Apr 29, 2022 at 8:25 history edited Denis Serre CC BY-SA 4.0
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Apr 29, 2022 at 8:23 comment added Denis Serre @NeilStrickland. Yes, that was actually my calculation, but when I typed, I forgot to symmetrize. Let me fix my answer.
Apr 29, 2022 at 8:21 comment added Neil Strickland I think that this is not quite right. There are six (not three) possible terms that are permutations of $(x-y)^2(w+z)^2$, and each should appear with coefficient $2$ (not $4$).
Apr 29, 2022 at 7:41 history answered Denis Serre CC BY-SA 4.0