I was going through a research paper which proves the existence of infinite family of rank $6$ elliptic curve over $\mathbb{Q}$ with invariant equal to $0$. Let $k$ be a field of characteristic zero. Let $p \in k[X]$ be a unit polynomial of degree $6$. Then there exists a unique unit polynomial $g \in k[X]$, of degree $2$, such that the polynomial $r = p-g^3$ is of degree $\leq 3$. Suppose the roots $x_1,\dotsc ,x_6$ of $p$ are in $k$. The curve $E$ with equation $r(x) + y^3 = 0$ contains the $6$ $k$-rational points $P_i = (r(x_i),g(x_i))$, $1\leq i\leq 6$. In the paper the author claims that an elliptic curve E, defined on $k(t)$, provided with $6$ $k(t)$-rational points. This curve is $k(t)$ isomorphic to the curve $y^2 = x^3 - 16D$, where $D$ is the discriminant of the polynomial $r$.

Reference of the paper: https://arxiv.org/abs/alg-geom/9206007

My question: Is the curve isomorphic to the curve $y^2 = x^3 - 16D$ or it should be the curve $y^2 = x^3 + 16D$ ?

I have tried to convert this cubic equation $y^3+r(x)=0$ into Weierstrass form and want to show that the Weierstrass form is same as $y^2 = x^3 - 16D$.

I have tried to convert this in sage:

Input in Sage:

R.<a,b,c, x, y, z> = QQ[]
cubic = x^3 + y^3 - a*x^2*z - b*x*z^2-c*z^3

WeierstrassForm(cubic, variables=[x,y,z])

Output:

(0, 1/4*a^2*b^2 - a^3*c + b^3 - 9/2*a*b*c - 27/4*c^2) = (0, D/4) = (0, 16D)

Where $D$ is discriminant of $x^3-ax^2-bx-c.$

Note: I know $y^3 + (x^3-n)=0$ is isomorphic to $y^2=x^3-432n^2$, this is well known, but note that $x^3-n$ has discriminant $-27n^2$. So it does not match with the claim made by author.

Please give some hints how to prove the explicit isomorphism.