In the GMB 7/2005, mentioned here, the solution to problem 25223 (published in GMB/1/2005, page 40) appears, which is the source of this post. (With nostalgia, I found on page 299 the name of my college colleague Constantin BUȘE...deceased in the meantime)
" 25223. To solve in $\mathbb{C}$ the equation : $3x^3-3x+1=0.$
[author : ] Gh. Szöllősy, Sighetul Marmației"
ANSWER
$x_1=\frac{2}{\sqrt{3}}\cos\frac{5\pi}{18}\;,\;x_2=\frac{2}{\sqrt{3}}\cos \frac{7\pi}{18}\;,\;x_3=\frac{2}{\sqrt{3}}\cos\frac{17\pi}{18}$
Solution
We are looking for solutions of the kind $x_t=a\cdot \cos t,\;a>0,\;t\in [0,\pi]$. Then
$3x_t^3-3x_t+1=0\Leftrightarrow 3a^3\cos^3x_t-3a\cos x_t+1=0 \tag{1}$
Because $\cos 3t=4\cos^3t-3\cos t$ we are looking to determine $a$ such that $\frac{3a^3}{4}=\frac{3a}{3}$ which leads to $a=\frac{2}{\sqrt{3}}.$ Thus we have
$\frac{2}{\sqrt{3}}=\frac{3a^3\cos^3x_t-3a\cos x_t}{4\cos^3 x_t-3\cos x_t}\underset{(1)}{=}\frac{-1}{\cos 3t}$
From $\cos 3t=-\frac{\sqrt{3}}{2}$ we obtain $3t \in \left \{\pm \frac{5\pi}{6}+2k\pi\right \}\cap [0,\pi]$, so $t\in \left \{\frac{5\pi}{18},\;\frac{7\pi}{18},\;\frac{17\pi}{18}\right\}$ hence the Answer.
$\blacksquare$
REMARK CiP Putting $x=\frac{2}{\sqrt{3}}\cdot y$ we see that $y$ are solutions of the equation
$3\cdot \frac{8}{3\sqrt{3}}y^3-3\cdot \frac{2}{\sqrt{3}}y+1=0\;\Leftrightarrow\;8y^3-6y+\sqrt{3}=0 \tag{2}$
that is $y_{1-3}\in \left \{\cos\frac{5\pi}{18},\;\cos\frac{7\pi}{18},\;\cos\frac{17\pi}{18}=-\cos\frac{\pi}{18}\right \}.$ We therefore have, from Vieta's Formulas :
$\cos\frac{\pi}{18}-\cos\frac{5\pi}{18}-\cos\frac{7\pi}{18}=0$
$\cos \frac{\pi}{18}\cdot \cos \frac{5\pi}{18}-\cos \frac{5\pi}{18}\cdot \cos \frac {7\pi}{18}+\cos \frac{\pi}{18}\cdot \cos \frac{7\pi}{18}=-\frac{3}{4}$
$\cos \frac{\pi}{18}\cdot \cos \frac{5\pi}{18} \cdot \cos \frac{7\pi}{18}=\frac{\sqrt{3}}{8}$
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