Sum of three cubes $x^3+y^3+z^3=2$ 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?


(Edit added after dburde's answer.)
P.S.: I know $$1214928^3+3480205^3-3528875^3=2$$
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.
 A: No, the solution $1214928^3 + 3480205^3 − 3528875^3=2$ is not of the form
$(6t^3+1,1-6t^3,-6t^2)$. 
A: If you're looking for all degree-3 polynomial solutions $x(t)=x_3t^3+x_2t^2+x_1t+x_0$, $y(t)=y_3t^3+y_2t^2+y_1t+y_0$ and $z(t)=z_3t^3+z_2t^2+z_1t+z_0$ of $x(t)^3+y(t)^3+z(t)^3=2$ you'll trivially find an infinity of them. The question of unicity must therefore cover the case of polynomials that are essentially the same as the well-known parametric solution, e.g. the solution with $x(t)=7+18t+18t^2+6t^3=1+6(t+1)^3$ is not different from $1+6t^3$.
For degree-3 solutions, it's easy to prove that the solution $1+6t^3, 1-6t^3, -6t^2$ is the only one with $(x_0,y_0,z_0)=(1,1,0)$.
More difficult is to answer (still for degree-3 solutions) if there is, or not, another solution with a triple $(x_0,y_0,z_0)$ not of the form $(1+6t_0^3, 1-6t_0^3, -6t_0^2)$.
Even more difficult, is to find out if there could be, or not, polynomial solutions of higher degree and essentially different from $1+6t^3, 1-6t^3, -6t^2$ (different in the sense that $t$ cannot be a polynomial $P(x))$.
For example, solutions with $X(t)=48t^6+144t^5-240t^3+144t-47$ are essentiellement the same as $x(t)=1+6t^3$ because $X(t)=1+6(2t^2+2t-2)^3$.
