Elliptic Curve Discriminant Here is my attempted approach to prove that the discriminant $\triangle = 4a^3+27b^2$ of an elliptic curve in the form of $y^2 = x^3 + ax +b$ is zero. I have a problem at the end which doesn't bring me to the expected conclusion. Below is the process, please let me know the error in the process. I try to approach this from the definition of non-singularity (part of the definition of an elliptic curve), which is equivalent to the statement: the equation $y^2 = x^3 + ax +b$ is differentiable everywhere on the graph. From that definition, I try to derive the derivative of $y$ in respect to $x$ through implicit differentiation:
\begin{align}
    y^2 &= x^3+ax+b\\ \label{ref1}
    2yy' &= 3x^2+a\\
    y' &= \frac{3x^2+a}{2y}
\end{align}
If the graph is singular, then $y'$ does not exist, in other words:
\begin{align}
    2y &= 0\\ 
    3x^2+a &\neq 0 
\end{align}
I replace $y = 0$ in the equation of elliptic curve which yields
\begin{equation}
   0 \ = \ x^3 + ax + b
\end{equation}
I apply Cardano's method, who realize that the form could be represented as
\begin{equation}
    (\alpha-\beta)^3 + 3\alpha\beta(\alpha-\beta) = \alpha^3 - \beta^3
\end{equation}
in which
\begin{align}
    \alpha\beta &= \frac{a}{3} \\
    \alpha^3 - \beta^3 &= -b 
\end{align}
By substituting $\alpha = \frac{a}{3\beta}$ in the second equation from above, I obtain
\begin{equation}
    (\frac{a}{3\beta})^3 - \beta^3 = -b
\end{equation}
I further simplify this by considering $\beta^3$ as a whole, i.e.
\begin{align}
    \frac{a^3}{27}-\beta^6 &= -b\beta^3\\
    (\beta^3)^2 - b\beta^3 -\frac{a^3}{27} &= 0
\end{align}
By quadratic formula,
\begin{equation}
    \beta^3 = \frac{b\pm \sqrt{b^2+\frac{4a^3}{27}}}{2} = \frac{b}{2} \pm \sqrt{\frac{b^2}{4}+\frac{a^3}{27}}
\end{equation}
From $\alpha^3 - \beta^3 = -b$, I get
\begin{equation}
    \alpha^3 = \beta^3 - b = -\frac{b}{2} \pm \sqrt{\frac{b^2}{4}+\frac{a^3}{27}}
\end{equation}
From $3x^2+a \neq 0$:
\begin{equation}
    3\left(\left(-\frac{b}{2} \pm \sqrt{\frac{b^2}{4}+\frac{a^3}{27}}\right)^\frac{1}{3} - \left(\frac{b}{2} \pm \sqrt{\frac{b^2}{4}+\frac{a^3}{27}}\right)^\frac{1}{3}\right)^2 + a \neq0
\end{equation}
The equation above is from $3(\alpha-\beta)^2+a \neq 0$. By De Moivre's Formula there are two other equations, $3\left(\alpha(\frac{-1}{2}+\frac{\sqrt{3}i}{2})-\beta(\frac{-1}{2}+\frac{\sqrt{3}i}{2})\right)^2+a \neq 0$ and $3\left(\alpha(\frac{-1}{2}+\frac{\sqrt{3}i}{2})^2-\beta(\frac{-1}{2}+\frac{\sqrt{3}i}{2})\right)^2+a \neq 0$. These result in several representations of b in terms of a and one integer solution, $a=0,b=0$ shouldn't exist at the same time, $b\neq\pm \frac{2ia^{3/2}}{3\sqrt{3}}$. These two meet my expectation, $\triangle = 0$. However, there are other four representations that do not meet my expectation $b \neq \pm \sqrt{\frac{a^3}{54}\pm\frac{5ia^3}{6\sqrt{3}}}$ and $b \neq \pm \sqrt{\frac{a^3}{54}\mp\frac{5ia^3}{6\sqrt{3}}}$. I also try to see what I could get from \begin{equation}
    3\left(\left(-\frac{b}{2}\right)^\frac{1}{3} - \left(\frac{b}{2}\right)^\frac{1}{3}\right)^2 + a \neq0
\end{equation}
which yields that $a=0,b=0$ shouldn't exist at the same time, or $b\neq\pm \frac{2ia^{3/2}}{3\sqrt{3}}$ these satisfy $\triangle = 0$ and also a weird pair of $b \neq \pm \frac{ia^{3/2}}{12\sqrt{3}}$, which doesn't even satisfy $\triangle = 0$. Could someone explain this?

When  $3x^2 + a \ne0$ and $2y = 0$, the curve is still nonsingular, its just that the slope of the tangent line is infinite.
The singularity occurs when
$$3x^2 + a = 0 \text{ and } 2y =0$$
simultaneously i.e.
$a = - 3x^2$ so $0 = x^3 + (-3x^2)x + b$
giving
$$ b = 2x^3$$
this is what leads to the equation
$$27b^2 + 4a^3 = 0$$
