The birational change of variables $(u,v) = (\frac{36+y}{6x},\frac{36-y}{6x})$ maps $u^3+v^3=1$ to $y^2 = x^3 - 432$ which has discriminant $-2^{12}\cdot 3^9$.
Using pari/gp we can compute the torsion subgroup:
? elltors(ellinit([0,0,0,0,-432]))
%1 = [3, [3], [[12, 36]]]
This says the torsion subgroup has order 3, is $\mathbf{Z}/3\mathbf{Z}$ and is generated by $(12,36)$ (which corresponds to $1^3+0^3=1^3$). The reason it has order 3 is because this also includes the projective solution $[0:0:1]$ of $X^3+Y^3=Z^3$.
Edit: By Nagell-Lutz one only needs to solve $y^2 = x^3 - 432$ in integers for $y=0$ and $y^2|2^{12}\cdot 3^9$ (which is a simple generate and test) to compute the elements of the torsion subgroup 'on paper'.
The group of rational points for this curve is then (by Mordell's Theorem) of the form $\mathbf{Z}^r \times \mathbf{Z}/3\mathbf{Z}$ where $r$ is the rank of the curve. If we can show the rank is 0 then this would prove fermats last theorem for $n = 3$.
How can it be shown directly the rank of this curve is 0?