# Something new in old question about sums of three polynomial cubes ?

An old problem asks whether or not the polynomial $$t \in \mathbb{Q}[t]$$ is a sum of three cubes, (of polynomials in $\mathbb{Q}[t]$).

Question: Something new known now ? Somebody has an idea of what to try (besides searching the literature available)?

It is well known and easy to write $t$ as a sum of four cubes. Also, certainly, $t$ cannot be a sum of two cubes.

More precisely: (a) $t$ is a sum of $4$ cubes in $\mathbb{Q}[t]$ since $$t =(t/6+1)^3- (t/6)^3 - (t/6)^3 +(t/6-1)^3$$ Some variants are

(b) $t$ is a sum $3$ cubes in $\mathbb{Q}[2^{1/3}][t]$ since $$t =(t/6+1)^3- {(2^{1/3} t/6)}^3 +(t/6-1)^3$$

(c) $t$ is a sum $3$ cubes in $\mathbb{Q}(t)$ since $$t =((t^3-1/27)^3+ {(t^2+ t/3)}^3 +(-t^3+t/3+1/27)^3)/(t^2+t/3+1/9)^3$$

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It is also well-known that, since $(t+1)^3-2t^3+(t-1)^3=6t$, there are solutions if you take your coefficients from an appropriate number field. –  Gerry Myerson Jan 24 '11 at 23:35
Your formula is the same formula that proves $4$ cubes over $\mathbb{Q}$ just change $t$ into $t/6.$ –  Luis H Gallardo Jan 25 '11 at 1:20
Edited using Gerry's idea above. @Gerry: Using your idea and the tangent-chord method you can build in the surface $t=x^3+y^3+z^3,$ more solutions $(x,y,z) \in {Q[2^{1/3}][t]}^3.$ I do not know if all these solutions are forced or not to be always outside of $Q[t]^3.$ –  Luis H Gallardo Jan 26 '11 at 1:16

@Luis: a search for "cubes" and "polynomials" in some combination, brings up some completely unrelated papers, and a bunch of papers co-authored by you (admittedly, not over $\mathbb{Q}.$ @Andrei: I don't disagree with you. I had a dim recollection of the foundational paper by Vaserstein, so figured if there were any progress, the paper would cite this. However, all the citing papers were co-authored by Luis, which, to me means that he is the world expert in the field. No insult was meant. –  Igor Rivin Jan 24 '11 at 23:04