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 $$ t =((t^3-1/27)^3+ {(t^2+ t/3)}^3 +(-t^3+t/3+1/27)^3)/(t^2+t/3+1/9)^3 $$
tangent-chordmethod 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 2011 at 1:16