Factor $x^3 + ax + b = (x-x_1) (x-x_2) (x-x_3)$. Translate $x$ by $x_1$ to make the first factor $x$. Then multiply $x$ by $x_2-x_1$ and $y$ by $(x_2-x_1)^{3/2}$ and divide by $(x_2-x_1)^3$ to get $y^2 = x (x-1) (x-L)$ with $L = (x_3-x_1)/(x_2-x_1)$. If $x^3+ax+b$ does not factor completely then $L$ must be in a field larger than the one used to define the curve.
– Noam D. ElkiesJun 19 '12 at 4:28

@Noam: This is an answer, not a comment.
– Martin BrandenburgJun 19 '12 at 9:39