This is in continuance to my earlier post : Deriving expression for general cubic equation solution. Also, the links are repeated for the first four pages of the book by Dickson, titled "Introduction to Theory of Algebraic Equations" are given as : page #1,$\,$page #2,$\,$ page #3, $\,$page #4
On page #4, the second question asks to show :
The cubic ($2$), i.e. the reduced cubic equation: $y^3+py+q=0$, with $$p = c_2-\frac{1}{3}c_1^2,\,\,\,\,\,\,\,\,\,\, q= -c_3+\frac{1}{3}c_1c_2 - \frac{2}{27}c_1^3$$ $$R = \frac{1}{4}q^2 + \frac{1}{27}p^3=\frac{1}{4}(-c_3+(\frac{1}{3}c_1c_2 - \frac{2}{27}c_1^3))^2 + \frac{1}{27}(c_2-\frac{1}{3}c_1^2)^3 = \frac{1}{4}(c_3^2+ (\frac{1}{9}c_1^2c_2^2 + \frac{4}{27^2}c_1^6 - \frac{4}{81}c_1^4c_2)-\frac{2}{3}c_1c_2c_3+\frac{4}{27}c_1^3c_3)+\frac{1}{27}(c_2^3+\frac{1}{27}c_1^6-c_2^2c_1^2+\frac{1}{3}c_2c_1^4) $$
$$=\frac{1}{4}c_3^2 + \frac{1}{36}c_1^2c_2^2+ \frac{1}{27^2}c_1^6-\frac{1}{81}c_1^4c_2 -\frac{1}{6}c_1c_2c_3 +\frac{1}{27}c_1^3c_3+\frac{1}{27}c_2^3+\frac{1}{27^2}c_1^6-\frac{1}{27}c_2^2c_1^2 +\frac{1}{81}c_2c_1^4$$
$$=\frac{1}{4}c_3^2+\frac{1}{36}c_1^2c_2^2+\frac{2}{27^2}c_1^6-\frac{1}{6}c_1c_2c_3+\frac{1}{27}c_1^3c_3+\frac{1}{27}c_2^3-\frac{1}{27}c_2^2c_1^3$$
has :
(i) one real root and two imaginary roots if $R\gt 0$,
(i) three real root, two of which are equal if $R= 0$,
(i) three real & distinct roots (irreducible case) if $R\lt 0$
This is generally not true for $p,q \in \mathbb{C}$. For example, consider the polynomial $y^3+i=0$. Then $p=0, q=i$, and $R=\frac14q^2+\frac1{27}p^3=-\frac14 <0$ but it has no real root.
However, this is true for $p,q \in \mathbb{R}$. If $p,q \in \mathbb{R}$, we have at least one real root and any complex roots comes in complex conjugate pair.
\begin{align}(y_1-y_2)^2(y_2-y_3)^2(y_3-y_1)^2&=-27q^2-4p^3\\&=-(27q^2+4p^3)\\&=-4(27)\left( \frac14q^2+\frac1{27}p^3\right)\\&=-4(27)R \end{align}
If $R>0$, then $(y_1-y_2)^2(y_2-y_3)^2(y_3-y_1)^2<0$, hence we must have an imaginary root. Hence we have one real root and two imaginary roots.
If $R=0$, then we must have at least a pair of repeating root as $(y_1-y_2)^2(y_2-y_3)^2(y_3-y_1)^2=0$
If $R<0$, then $(y_1-y_2)^2(y_2-y_3)^2(y_3-y_1)^2>0$. Let's rule out the possibility that we have any complex root. Suppose $y_1 \in \mathbb{R}, y_2, y_3$ are complex conjugate. Then
\begin{align}(y_1-y_2)^2(y_2-y_3)^2(y_3-y_1)^2&=(y_1-y_2)^2(y_2-\bar{y_2})^2(\bar{y_2}-y_1)^2\\&=|y_1-y_2|^4(2\Im(y_2))i)^2\\&=-4 \Im(y_2)^2|y_1-y_2|^4 \le 0\end{align} which is a contradicition.
Hence if $R<0$, every root is distinct.