Making an elliptic curve out of a cubic polynomial made a cube, or $ax^3+bx^2+cx+d = y^3$ What is the transformation such that a general cubic polynomial to be made a cube,
$$ax^3+bx^2+cx+d = y^3\tag{1}$$
can be transformed to Weierstrass form,
$$x^3+Ax+B = t^2\tag{2}$$
(The special case $b = c = 0$ is easier.) Given an initial rational point, I know a method how to find subsequent ones, but it would be nice to know the general transformation. For example,
$$3x^3+9x^2+15x+9 = y^3\tag{3}$$
I find that,
$$x_1 = 3$$
$$x_2 = -1839/1871$$
$$x_3 = -13898941449153/12222218425537$$
and so on. (I may have skipped some points.) But how do you transform $(3)$ to $(2)$?
Postscript (After Jyrki's answer)
For those interested, the cubic $(3)$ in two-variable form is equivalent to,
$$3p^3+9p^2q+15pq^2+9q^3 =\; p^3 + (p+q)^3+(p+2q)^3 = t^3$$
or three cubes (not necessarily positive) in arithmetic progrees. Thus, $p,q = 3,1$ gives the well-known,
$$3^3+4^3+5^3 = 6^3$$
and $p,q = -1839, 1871$ yields,
$$(-1839)^3+(-1839+1871)^3+(-1839+2\cdot1871)^3 = 876^3$$

General algorithm for transforming a cubic into Weierstrass form, with several worked examples, can be found in Section 1.4 of Ian Connell's Elliptic Curve Handbook 
