Answer: Using Vieta’s formulae for the cubic \(x^{3}+px^{2}+qx+r=0\) with roots \(\frac{\beta}{k},\,\beta,\,k\beta\):

1. The sum of the roots is \(-p\), so \(\frac{\beta}{k}+\beta+k\beta=-p\).
2. The sum of the products of roots in pairs is \(q\), so \(\frac{\beta^{2}}{k}+\beta^{2}+k\beta^{2}=q\).
3. The product of the roots is \(-r\), so \(\beta^{3}=-r\).

From the first two equations, factor out the common bracket \(\frac{k^{2}+k+1}{k}\): \(\beta\left(\frac{k^{2}+k+1}{k}\right)=-p\) and \(\beta^{2}\left(\frac{k^{2}+k+1}{k}\right)=q\). Dividing the second by the first gives \(-\beta=\frac{q}{p}\), hence \(\beta=-\frac{q}{p}\).

Then \(\beta^{3}=-r\) gives \(-\frac{q^{3}}{p^{3}}=-r\), so \(rp^{3}=q^{3}\).

Let the roots be \(\dfrac{\beta}{k},\,\beta,\,k\beta\).

By Vieta’s formulae for \(x^{3}+px^{2}+qx+r=0\):

• sum of roots \(= -p\),
• sum of pairwise products \(= q\),
• product of roots \(= -r\).

So

\(\dfrac{\beta}{k}+\beta+k\beta=-p\).

Multiply through by \(k\):

\(\beta+\beta k+\beta k^{2}=-pk\),

hence

\(\beta\left(\dfrac{k^{2}+k+1}{k}\right)=-p\).

Next, the pairwise products are

\(\dfrac{\beta}{k}\cdot\beta=\dfrac{\beta^{2}}{k},\quad \dfrac{\beta}{k}\cdot k\beta=\beta^{2},\quad \beta\cdot k\beta=k\beta^{2}\),

so

\(\dfrac{\beta^{2}}{k}+\beta^{2}+k\beta^{2}=q\),

which gives

\(\beta^{2}\left(\dfrac{k^{2}+k+1}{k}\right)=q\).

Now divide this equation by the earlier one:

\(\dfrac{\beta^{2}\left(\dfrac{k^{2}+k+1}{k}\right)}{\beta\left(\dfrac{k^{2}+k+1}{k}\right)}=\dfrac{q}{-p}\),

so \(\beta=-\dfrac{q}{p}\).

For the product of the roots,

\(\dfrac{\beta}{k}\cdot\beta\cdot k\beta=\beta^{3}=-r\).

Substitute \(\beta=-\dfrac{q}{p}\):

\(\left(-\dfrac{q}{p}\right)^{3}=-r\),

so

\(-\dfrac{q^{3}}{p^{3}}=-r\).

Therefore \(rp^{3}=q^{3}\).