Answer: Let the repeated root be \(\alpha\), so the roots are \(\alpha,\alpha,\beta\).
By Vieta’s relations for \(x^{3}+px+q=0\),
\(2\alpha+\beta=0\), \(\alpha^{2}+2\alpha\beta=p\), and \(\alpha^{2}\beta=-q\).
From \(2\alpha+\beta=0\), we get \(\beta=-2\alpha\).
Substitute into \(\alpha^{2}+2\alpha\beta=p\):
\(p=\alpha^{2}+2\alpha(-2\alpha)=\alpha^{2}-4\alpha^{2}=-3\alpha^{2}.\)
Substitute into \(\alpha^{2}\beta=-q\):
\(\alpha^{2}(-2\alpha)=-q\), so \(q=2\alpha^{3}.\)
Now eliminate \(\alpha\). Since \(p=-3\alpha^{2}\), we have \(\alpha^{2}=-\frac{p}{3}\), so
\(\alpha^{6}=\left(-\frac{p}{3}\right)^{3}=-\frac{p^{3}}{27}.\)
Also, \(q=2\alpha^{3}\), hence \(q^{2}=4\alpha^{6}=4\left(-\frac{p^{3}}{27}\right)=-\frac{4p^{3}}{27}.\)
Therefore
\(27q^{2}=-4p^{3}\),
so
\(4p^{3}+27q^{2}=0\).
Suppose the repeated root is \(\alpha\). Then the three roots are \(\alpha,\alpha,\beta\).
For the monic cubic \(x^{3}+px+q=0\), Vieta’s relations give
\(\alpha+\alpha+\beta=0\),
\(\alpha\alpha+\alpha\beta+\alpha\beta=p\),
and
\(\alpha\cdot\alpha\cdot\beta=-q\).
So
\(2\alpha+\beta=0\), \(\alpha^{2}+2\alpha\beta=p\), \(\alpha^{2}\beta=-q\).
From \(2\alpha+\beta=0\),
\(\beta=-2\alpha\).
Substitute this into the expression for \(p\):
\(p=\alpha^{2}+2\alpha(-2\alpha)=\alpha^{2}-4\alpha^{2}=-3\alpha^{2}.\)
Substitute \(\beta=-2\alpha\) into the product relation:
\(\alpha^{2}(-2\alpha)=-q\),
so \(q=2\alpha^{3}\).
Now square the expression for \(q\):
\(q^{2}=4\alpha^{6}.\)
Since \(p=-3\alpha^{2}\), we have
\(\alpha^{2}=-\frac{p}{3}\), hence
\(\alpha^{6}=\left(-\frac{p}{3}\right)^{3}=-\frac{p^{3}}{27}.\)
Therefore
\(q^{2}=4\left(-\frac{p^{3}}{27}\right)=-\frac{4p^{3}}{27}.\)
Multiplying by \(27\) gives
\(27q^{2}=-4p^{3}.\)
Rearranging,
\(4p^{3}+27q^{2}=0\), as required.
