<s>[INST] Loney: If $\alpha$, $\beta$, $\gamma$ are the roots of $x^3 + px^2 + qx + p = 0$, then $\tan^{-1}\alpha + \tan^{-1}\beta + \tan^{-1}\gamma = n\pi$ 
If $\alpha, \beta, \gamma$ are the roots of the equation $$x^3 + px^2 + qx + p = 0,$$
  prove that $$\tan^{-1}\left(\alpha\right) + \tan^{-1}\left(\beta\right) + \tan^{-1}\left(\gamma\right) = n\pi$$ 
  except in one particular case. 

This question is from S. L. Loney's 'Plane Trigonometry' page 327 q13. 
It may be useful to note that this section utilises $$\tan\left(\alpha + \beta + \gamma + ...\right) = \frac{s_1 - s_3 + s_5}{1 - s_2 + s_4 - \cdots}$$
where
$s_1 =$ the sum of the tangents of the separate angles,
$s_2 =$ the sum of the tangents taken two at a time,
$s_3 =$ the sum of the tangents taken three at a time, and so on.
I do not know where I should start with this question. 
 [/INST] By Vieta’s formula we have
$$-abc=p$$
$$-(a+b+c)=p$$
$$ab+ac+bc=q$$
By using the formula
$$\tan^{-1}u+\tan^{-1}v=\tan^{-1}\frac{u+v}{1-uv}\mod{\pi}\qquad{uv\ne 1}$$, we have
$$
\begin{align}
\tan^{-1}a+ \tan^{-1} b+ \tan^{-1} c
&= \tan^{-1} \frac{a+b}{1-ab}+ \tan^{-1} c \\
&=\tan^{-1}\frac
{\frac{a+b}{1-ab}+c}
{1-\frac{(a+b)c}{1-ab}} \\
&=\tan^{-1}\frac
{\frac{a+b+c-abc}{1-ab}}
{\frac{1-ab-ac-bc}{1-ab}} \\
&=\tan^{-1}\frac{(a+b+c)-abc}{1-(ab+ac+bc)} \\
&=\tan^{-1}\frac{-p+p}{1-q}\\
&=0 \mod{\pi}
\end{align}
$$
We have assumed that $ab\ne 1$ and $q\ne 1$.

It can be shown that
$$ab\ne 1\implies q\ne 1$$
Substitute $ab=1$ into the first three Vieta’s equations,
$$-c=p$$
$$a+b+c=-p\implies a+b=-p-c=0$$
$$1+(a+b)c=q\implies q=1$$
Therefore, $q\ne1$ is the stronger condition, and $q=1$ should be the only exception that the question expects you to give.
</s>