Problem source

The cubic equation 
\[
x^{3}-px^{2}+qx-r=0
\]
has roots $a,b$ and $c$. Express $p,q$ and $r$ in terms of $a,b$ and $c$. 
\begin{questionparts}
\item If $p=0$ and two of the roots are equal to each other, show that 
\[
4q^{3}+27r^{2}=0.
\]
\item Show that, if two of the roots of the original equation are equal to each other, then 
\[
4\left(q-\frac{p^{2}}{3}\right)^{3}+27\left(\frac{2p^{3}}{27}-\frac{pq}{3}+r\right)^{2}=0.
\]
\end{questionparts}

Solution source

$p = a+b+c, q = ab+bc+ca, r = abc$

\begin{questionparts}
\item Suppose two roots are equal to each other, this means that one of the roots is also a root of the derivative. ie
\begin{align*}
&& 0 &= x^3+qx - r \\
&& 0 &= 3x^2+q
\end{align*}

have a common root, but this root must satisfy $x^2 = -\frac{q}{3}$. Then

\begin{align*}
&&0 &= x^3 + qx - r \\
&&&= x^3 -3x^3 - r \\
&&&= -2x^3 -r \\
\Rightarrow && r^2 &= 4x^6 \\
&&&= 4 \left ( -\frac{q}{3}\right)^3 \\
\Rightarrow && 0 &= 27r^2+4q^3
\end{align*}

\item Consider $x = z + \frac{p}{3}$, then the equation is:

\begin{align*}
x^{3}-px^{2}+qx-r &= (z + \frac{p}{3})^3 - p(z + \frac{p}{3})^2 + q(z + \frac{p}{3}) - r \\
&= z^3 + pz^2 + \frac{p^2}{3}z + \frac{p^3}{27} - \\
&\quad -pz^2-\frac{2p^2}{3}z-\frac{p^3}{9} + \\
&\quad\quad qz + \frac{pq}{3} - r \\
&= z^3+\left (\frac{p^2}{3}-\frac{2p^2}{3}+q \right)z + \left (\frac{p^3}{27}-\frac{p^3}{9}+\frac{pq}{3}-r \right) \\
&= z^3+\left (-\frac{p^2}{3}+q \right)z + \left (-\frac{2p^3}{27}+\frac{pq}{3}-r \right) \\
\end{align*}

Since this equation must also have repeated roots we must have:

\begin{align*}
4\left (-\frac{p^2}{3}+q \right)^3 + 27 \left (-\frac{2p^3}{27}+\frac{pq}{3}-r \right)^2 = 0
\end{align*}

which is exactly our desired result

\end{questionparts}