Problem source

Consider the cubic equation
\[
x^3-px^2+qx-r=0\;,
\]
where $p\ne0$ and $r\ne 0$.
\begin{questionparts}
\item If the three roots can be written in the form $ak^{-1}$, $a$ and $ak$ for some constants $a$ and $k$, show that one root is $q/p$ and that $q^3 -rp^3=0\;.$
\item If $r=q^3/p^3\;$, show that $q/p$ is a root and that the product of the other two roots is $(q/p)^2$. Deduce that the roots are in geometric progression.
\item Find a necessary and sufficient condition involving $p$, $q$ and $r$ for the  roots to be  in  arithmetic progression.
\end{questionparts}

Solution source

\begin{questionparts}
\item If the roots are $ak^{-1}, a, ak$ then we must have that $p = a(k^{-1}+1+k)$, $q = a^2(k^{-1}+k+1)$ and $r = a^3$, therefore $a = \frac{q}{p}$ (ie one of the roots is $\frac q p$ and $r = \left ( \frac{q}{p} \right)^3 \Rightarrow q^3 =rp^3 \Rightarrow q^3-rp^3 = 0$

\item Suppose $r = q^3/p^3$ then

$\left (\frac{q}{p} \right)^3 - p\left (\frac{q}{p} \right)^2+q\left (\frac{q}{p} \right) - r = \frac{--pq^2+pq^2}{p^2} =0 $, therefore $q/p$ is a root by the factor theorem. We must also have the product of the three roots is $q^3/p^3$ but one of the roots is $q/p$ therefore the product of the other two roots is $q^2/p^2$, but the condition $ac = b^2$ is precisely the condition that $a,b,c$ is a geometric progression.

\item If the three roots are $a-d, a, a+d$ then $p = 3a$, $q = a^2-da+a^2+da+a^2-d^2 = 3a^2-d^2$, $r = a(a^2-q^2)$, therefore $\frac{p}{3}\left (q-\frac{2p^2}9 \right) = r$

Similarly, suppose $\frac{p}{3}$ is a root, then the other two roots must sum to twice this and therefore they are in arithmetic progression. The condition $\frac{p}{3}$ is a root is equivalent to:

$\frac{p^3}{27} - \frac{p^3}{9} + \frac{qp}{3} - r = 0$, ie exactly $\frac{p}{3}\left (q-\frac{2p^2}9 \right) = r$, therefore this condition is both necessary and sufficient.
\end{questionparts}