short weierstrass form for cubic I have some question about the derivation of the short Weierstrass form. In https://www.staff.uni-mainz.de/dfesti/EllipticCurvesNotes.pdf this note, I follow the derivation till the point $y^2 = β_0x^3 + β_1x^2 + β_2x + β_3$, but then it says that using the transformation $x' = x + β_1/3, y'=y$, we then arrive at $y^2 = x^3 + ax + b$. But my question is how it turns the coefficient $\beta_0$ into $1$? One method I come up with is to let $y^2 = β_0(x^3 + \gamma_1x^2 + \gamma_2x + \gamma_3)$, then applying $x' = x + \gamma_1/3, y'=y$, we get $y^2 = \beta_0(x^3 + ax + b)$ then change by a scalar $x' = \beta_0^{1/3}x, y'=y$, we get the form $y^2 = x^3 + ax + b$. but the problem is the final step would need to assume that $\beta_0$ has a cube root, which is not true from general rational numbers. Does anyone knows how to solve this? for a specific example, how could one turn the equation $y^2 = 5x^3+x+1$ into the short Weierstrass form?

You can get rid of the $\beta_0$ by multiplying both sides by $\beta_0^2$ and then letting $u = \beta_0 x$ and $v = \beta_0 y$. So in your example $y^2 = 5x^3+x+1$, we have
\begin{align*}
y^2 &= 5x^3+x+1\\
5^2 y^2 &= 5^3 x^3 + 5^2 x + 5^2\\
(5y)^2 &= (5x)^3 + 5(5x) + 25\\
v^2 &= u^3 + 5 u + 25 \, .
\end{align*}
