- Given that the roots of the equation \(x^3+px^2+qx+r=0\) are three consecutive terms of an arithmetic progression, show that \[2p^3 + 27r = 9pq.\]

Suppose the cubic equation \(ax^3+bx^2+cx+d=0\) has the three roots \(x_1\), \(x_2\) and \(x_3\). We can now write \[ax^3+bx^2+cx+d=a(x-x_1)(x-x_2)(x-x_3).\]

If we multiply out the right hand side, we have

\[ax^3+bx^2+cx+d=ax^3-a(x_1+x_2+x_3)x^2+ a(x_1x_2+x_3x_1+x_2x_3)x - ax_1x_2x_3.\]

This holds for all values of \(x\), so corresponding coefficients must be equal. This means \[\begin{align*} x_1+x_2+x_3 = -\dfrac{b}{a},\\ x_1x_2+x_3x_1+x_2x_3 = \dfrac{c}{a},\\ x_1x_2x_3 = -\dfrac{d}{a}. \end{align*}\]These equations connecting the coefficients of an equation with its roots are extremely useful.

So \(m^2 = 3n^2-q\), and so \(n^3-n(3n^2-q)=-r,\) or \(2n^3=r+nq\).

Substituting in for \(n\), we reach \(2\left(\dfrac{-p}{3}\right)^3=r+q\dfrac{-p}{3}\). Now multiplying out gives \[2p^3 + 27r = 9pq\] as required.

- Given that the roots of the equation \(x^3+px^2+qx+r=0\) are three consecutive terms of a geometric progression, find a condition that \(p\), \(q\) and \(r\) must satisfy.

Let’s say this time that our roots take the values \(n\), \(nm\) and \(nm^2\).

Using our work above on roots and coefficients for a second time, we have the three equations \[\begin{align*} n+nm+nm^2 & = -p, \\ n^2m+n^2m^2+n^2m^3 & = q, \\ n^3m^3 & = -r. \end{align*}\]The first equation tells us that \(1+m+m^2 = \dfrac{-p}{n}\).

The second says \(n^2m(1+m+m^2) = q\), so on substituting in for \(1+m+m^2\), we have \[n^2m\dfrac{-p}{n} = q,\] and so \(nm = \dfrac{-q}{p}\), and the third equation then gives us \[\left(\dfrac{-q}{p}\right)^3=-r,\] which is \[rp^3=q^3.\]

This is the required condition.