Solve the following cubic equation:

$$
x^{3}-7 x+7=0 .
$$
Minthogy

$$
\left(\frac{7}{2}\right)^{2}+\left(-\frac{7}{3}\right)^{3}=-\frac{49}{108}<0
$$

the equation has three real roots, which are equal to the following values:

$$
\begin{gathered}
x_{1}=2 \sqrt[3]{r} \sin \frac{\alpha}{3} \\
x_{2}=2 \sqrt[3]{r} \sin \left(\frac{\alpha}{3}+120^{\circ}\right) \\
-x_{3}=2 \sqrt[3]{r} \sin \left(\frac{\alpha}{3}+60^{\circ}\right)
\end{gathered}
$$

where

$$
\begin{gathered}
\sqrt[3]{r}=\sqrt{\frac{7}{3}}, \quad \sin \alpha=\frac{\frac{7}{2}}{\frac{7}{3} \sqrt{\frac{7}{3}}}=\frac{3}{2 \sqrt{\frac{7}{3}}} \\
2 \sqrt[3]{r}=\sqrt{\frac{28}{3}}=k
\end{gathered}
$$

The calculation process is as follows:

$$
\begin{aligned}
& \log 28=1,44716 \\
& \underline{\log 3=0,47712} \\
& 0,97004 \\
& \log k=0,48502 \\
& 9,64749-10 \\
& \log \sin \frac{\alpha}{3}=\frac{3}{9,64752-10} \\
& \log \sin \left(\frac{\alpha}{3}+120^{\circ}\right)= \\
& \log x_{1}=0,48502+ \\
& +9,64752-10 \\
& 0,13254 \\
& \underline{226} \\
& \log \sin \left(\frac{\alpha}{3}+60^{\circ}\right)=9,99913-10 \\
& \log x_{2}=0,48502+ \\
& +9,74349-10 \\
& 0,22841 \\
& \underline{840} \\
& x_{1}=1,3569 \\
& x_{2}=1,6920 \\
& 1,692 x_{3}=3,0489 \\
& 0 \overline{x_{1}+x_{2}+x_{3}=0}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_104598cd668cbf0d15cfg-2.jpg?height=52&width=44&top_left_y=2176&top_left_x=607)

$$
\begin{array}{r}
\log \left(-x_{3}\right)=0,48502+ \\
+9,99913-10 \\
\hline 0,48415 \\
3,048 \\
9
\end{array}
$$

László Makkai, fógymn. VIII. o. t. Nagy-Enyed).