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're right, there's a step missing here.
Starting from $$y^2 = \beta_0x^3 + \beta_1x^2 + \beta_2+ \beta_3$$ we can get rid of the (nonzero) $\beta_0$ as follows: send $y\mapsto \beta_0^2y$ and $x\mapsto \beta_0x$, which gives us $$\beta_0^4y^2 = \beta_0^4x^3 + \beta_0^2\beta_1x^2 + \beta_0\beta_2x+ \beta_3$$ and after dividing by $\beta_0^4$ we get $$y^2=x^3+\frac{\beta_1}{\beta_0^2}x^2+\frac{\beta_2}{\beta_0^3}x+\frac{\beta_3}{\beta_0^4}$$ so we may assume we have an equation of the form $y^2=x^3+\gamma_1x^2+\gamma_2x+\gamma_3$.
Now to get rid of $\gamma_1$, make the substitution $x=x-\frac{\gamma_1}{3}$. This sends $$x^3+\gamma_1x^2+\gamma_2x+\gamma_3 \mapsto (x-\frac{\gamma_1}{3})^3+\gamma_1(x-\frac{\gamma_1}{3})^2+\gamma_2(x-\frac{\gamma_1}{3})+\gamma_3,$$ and the right-hand side is equal to $$x^3+(\gamma_2-\frac{\gamma_1^2}{3})x+(\gamma_3-\frac{\gamma_1\gamma_2}{3}+\frac{2\gamma_1^3}{27})$$ which is the desired form.
