$• \ \text{a)}$ \begin{matrix} \tan{3\alpha} &=& \tan{(2\alpha + \alpha)} &=& \Large{\frac{\tan{2\alpha} \ + \ \tan{\alpha} }{1- \tan{2\alpha} \ . \ \tan{\alpha}}} &=& \Large{ \frac{ (\frac{2\tan{\alpha}}{1-\tan^2{\alpha}}) \ + \ \tan{\alpha}}{ 1 \ - \ (\frac{2\tan{\alpha}}{1-\tan^2{\alpha}}) \ . \ \tan{\alpha}}} &=& \Large{\frac{ 3\tan{3\alpha} \ - \ \tan^3{\alpha}}{ 1 \ - \ 3\tan^2{\alpha}} } \end{matrix} $• \ \text{b)}$ \begin{matrix} \tan^3{\alpha} - 3m\tan^2{\alpha} - 3\tan{\alpha} + m = 0 &\Rightarrow& 3\tan{\alpha} - \tan^3{\alpha} = m (1 - 3\tan^2{\alpha}) &\Rightarrow& {\Large{\frac{ 3\tan{3\alpha} \ - \ \tan^3{\alpha}}{ 1 \ - \ 3\tan^2{\alpha}} } } = m \end{matrix}Assim, \begin{matrix} \tan{3\alpha} = m &\Rightarrow& \alpha = \large{\frac{\arctan{m}}{3}} &\therefore& \fbox{$ x = \tan \large{(\frac{\arctan{m}}{3})} $} \end{matrix}