Answer: Let the roots be \(\alpha\), \(\alpha^{-1}\) and \(\beta\). Since the coefficient of \(x^2\) is zero, their sum is
\(\alpha+\alpha^{-1}+\beta=0\).
Also, for the monic cubic \(x^3+px+q=0\), the product of the roots is \(-q\). Hence
\(\alpha\cdot \alpha^{-1}\cdot \beta=\beta=-q\).
The sum of the products of the roots taken two at a time is \(p\), so
\(\alpha\alpha^{-1}+\alpha\beta+\alpha^{-1}\beta=1+\beta(\alpha+\alpha^{-1})=p\).
From \(\alpha+\alpha^{-1}=-\beta\), this gives
\(1-\beta^2=p\).
Using \(\beta=-q\), we obtain
\(p=1-q^2\).
Therefore \(p+q^2=1\).
Let the three roots of \(x^3+px+q=0\) be \(\alpha\), \(\alpha^{-1}\) and \(\beta\), where \(q\neq 0\). The condition that one root is the reciprocal of another allows us to label two roots in this way.
By Vieta's formula for the coefficient of \(x^2\), the sum of the roots is zero:
\(\alpha+\alpha^{-1}+\beta=0\). Hence \(\alpha+\alpha^{-1}=-\beta\).
For a monic cubic \(x^3+px+q\), the product of the roots is \(-q\). Therefore
\(\alpha\cdot \alpha^{-1}\cdot \beta=\beta=-q\).
Next, the sum of the products of the roots taken two at a time is the coefficient of \(x\), namely \(p\):
\(\alpha\alpha^{-1}+\alpha\beta+\alpha^{-1}\beta=p\).
Since \(\alpha\alpha^{-1}=1\), this becomes
\(1+\beta(\alpha+\alpha^{-1})=p\).
Using \(\alpha+\alpha^{-1}=-\beta\), we get
\(1+\beta(-\beta)=p\), so
\(1-\beta^2=p\).
Now substitute \(\beta=-q\):
\(p=1-(-q)^2=1-q^2\).
Therefore
\(p+q^2=1\).
