# sum of three cubes and parametric solutions

The first paragraph in the following link asserts that the equation $x^3+y^3+z^3=2$ has finite many parametric solutions over $\mathbb{Q}$. In other words, there are finite many polynomial triples $(x(t),y(t),z(t))$ with $x(t),y(t),z(t)\in\mathbb{Q}[t]$ satisfying the equation $x^3(t)+y^3(t)+z^3(t)=2$.

Question: What might be the exact evidence for such an assertion? Is it possible that $(1+6t^3,1-6t^3,-6t^2)$ is the unique solution to this equation?

P.S.: I know $$3480205^3+1214928^3-3528875^3=2$$ found by D.R.Heath-Brown

and $$37404275617^3-25282289375^3-33071554596^3=2$$

by D.J.Bernstein.

But I think this one

$$3737830626090^3+1490220318001^3-3815176160999^3=2$$

is not reported before.

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No, the solution $1214928^3 + 3480205^3 − 3528875^3=2$ is not of the form $(6t^3+1,1-6t^3,-6t^2)$.
This was found by D. R. Heath-Brown (Oxford) in $1993$, I believe... – dburde Apr 5 '13 at 15:05
@dburde: It is my fault that I did not fully articulate my question. What I want to express is: "Is it possible that $(1+6t^3,1−6t^3,−6t^2)$ is the unique parametric solution to this equation? – y zh Apr 6 '13 at 2:44