三次方程式の判別式の導出

\begin{align}
f(X)&=X^3+pX+q=(X-\alpha)(X-\beta)(X-\gamma)\\
f'(\alpha)&=(\alpha-\beta)(\alpha-\gamma)\\
f'(\beta)&=(\beta-\alpha)(\beta-\gamma)\\
f'(\gamma)&=(\gamma-\alpha)(\gamma-\beta)\\
D&=(\alpha-\beta)^2(\beta-\gamma)^2(\gamma-\alpha)^2\\
&=-f'(\alpha)f'(\beta)f'(\gamma)\\
&=-(3\alpha^2+p)(3\beta^2+p)(3\gamma^2+p)\\
&=-27\left(\alpha^2+\frac{p}{3}\right)
\left(\beta^2+\frac{p}{3}\right)
\left(\gamma^2+\frac{p}{3}\right)\\
&=-27\left(\alpha+i\sqrt{\frac{p}{3}}\right)\left(\alpha-i\sqrt{\frac{p}{3}}\right)
\left(\beta+i\sqrt{\frac{p}{3}}\right)\left(\beta-i\sqrt{\frac{p}{3}}\right)
\left(\gamma+i\sqrt{\frac{p}{3}}\right)\left(\gamma-i\sqrt{\frac{p}{3}}\right)\\
&=-27f\left(i\sqrt{\frac{p}{3}}\right)f\left(-i\sqrt{\frac{p}{3}}\right)\\
&=-27\left(-i\frac{p}{3}\sqrt{\frac{p}{3}}+ip\sqrt{\frac{p}{3}}+q\right)
\left(i\frac{p}{3}\sqrt{\frac{p}{3}}-ip\sqrt{\frac{p}{3}}+q\right)\\
&=-27\left(2i\frac{p}{3}\sqrt{\frac{p}{3}}+q\right)
\left(-2i\frac{p}{3}\sqrt{\frac{p}{3}}+q\right)\\
&=-27\left(\frac{4}{27}p^3+q^2\right)\\
&=-4p^3-27q^2
\end{align}