Determine the third-degree equation admitting the roots: $2 \sin \frac{2 \pi}{9}, 2 \sin \frac{8 \pi}{9}, 2 \sin \frac{14 \pi}{9}$.
The three reals $\sin \frac{2 \pi}{9}, \sin \frac{8 \pi}{9}, \sin \frac{14 \pi}{9}$ are the three imaginary parts of $x, x j, x j^{2}$ where $x$ is any of the three roots of the equation $X^{3}=j . x$ is here an algebraic number, defined solely by the equation of which it is a root, and not by its value as a complex number because the calculations will be identical whether $x=\cos \frac{2 \pi}{9}+i \sin \frac{2 \pi}{9}, x=\cos \frac{8 \pi}{9}+i \sin \frac{8 \pi}{9}$ or $x=\cos \frac{14 \pi}{9}+i \sin \frac{14 \pi}{9}$. In all three cases, the three numbers $2 \sin \frac{2 \pi}{9}, 2 \sin \frac{8 \pi}{9}, 2 \sin \frac{14 \pi}{9}$ are, up to permutation, the three algebraic numbers $i(\bar{x}-x), i(\overline{x j}-x j), i\left(\overline{x j^{2}}-x j^{2}\right)$. The symmetric polynomials of these three roots are:

$$
i(\bar{x}-x)+i(\overline{x j}-x j)+i\left(\overline{x j^{2}}-x j^{2}\right)=0
$$

because $1+j+j^{2}=0$. This same relation implies: $(\bar{x}-x)(\overline{x j}-x j)=\overline{x^{2} j}+1+x^{2} j$ so that:

$$
\begin{aligned}
-(\bar{x}-x)(\overline{x j}-x j)-(\overline{x j}-x j)\left(\overline{x j^{2}}-x j^{2}\right)-\left(\overline{x j^{2}}-x j^{2}\right)(\bar{x}-x) & = \\
-\left(\overline{x^{2} j}+1+x^{2} j\right)-\left(\overline{x^{2}}+1+x^{2}\right)-\left(\overline{x^{2} j^{2}}+1+x^{2} j^{2}\right) & =-3
\end{aligned}
$$

and

$$
\begin{aligned}
-i(\bar{x}-x)(\overline{x j}-x j)\left(\overline{x j^{2}}-x j^{2}\right) & =-i(\bar{x}-x)\left(\overline{x^{2}}+1+x^{2}\right) \\
& \left.=-i\left[\overline{x^{3}}-\bar{x}\right)+(\bar{x}-x)+\left(x-x^{3}\right)\right] \\
& =-i(\bar{j}-j)=-\sqrt{3}
\end{aligned}
$$

$2 \sin \frac{2 \pi}{9}, 2 \sin \frac{8 \pi}{9}, 2 \sin \frac{14 \pi}{9}$. are therefore the three roots of the equation: $X^{3}-3 X+\sqrt{3}=0$, which can be easily verified by solving this equation using Cardano's method.

The important point is that only $j$ and the relations satisfied at the level of $j$, namely: $j^{3}=1,1+j+j^{2}=0$ and $\bar{j}-j=-i \sqrt{3}$, as well as the relation $x^{3}=j$ (thus $x \bar{x}=1$) defining, at the higher level, $x$ as an algebraic number, were used in these calculations.

## The Heptagon, a Regular Polygon with Seven Sides

What about the heptagon, a regular polygon with seven sides?

Constructing a regular heptagon, even if it is not possible with a ruler and compass, amounts to solving the equation: $X^{7}-1=0$ or $(X-1)\left(X^{6}+X^{5}+X^{4}+X^{3}+X^{2}+X+1\right)=0$. One way to solve this equation, apart from the obvious root 1, is to make the change of variable: $Y=X+\frac{1}{X}$. Indeed,

$$
\begin{aligned}
Y^{3}-2 Y & =X^{3}+X+\frac{1}{X}+\frac{1}{X^{3}} \\
Y^{2}-1 & =X^{2}+1+\frac{1}{X^{2}}
\end{aligned}
$$

so that:

$$
\frac{1}{X^{3}}\left(X^{6}+X^{5}+X^{4}+X^{3}+X^{2}+X+1\right)=Y^{3}+Y^{2}-2 Y-1
$$

This last equation can be solved using Cardano's method, provided that "the" solution involves cube roots of complex numbers. If $Z=3 Y+1$, $Z^{3}-21 Z-7=27\left(Y^{3}+Y^{2}-2 Y-1\right)$, so it suffices to find $u$ and $v$ such that $u+v=Z$ is a root of this last equation, with $u v=7$, which gives: $u^{3}+v^{3}-7=0$ or $u^{6}-7 u^{3}+7^{3}=0$, thus $u^{3}=\frac{7+21 i \sqrt{3}}{2}=7(3 j+2)$ and $v^{3}=\frac{7-21 i \sqrt{3}}{2}=7\left(3 j^{2}+2\right)$. If $u$ is any of the cube roots of $7(3 j+2)$ and $v$ the corresponding cube root (such that $u v=7$) of $7\left(3 j^{2}+2\right)$, then the three roots $y_{1}, y_{2}, y_{3}$ of $Y^{3}+Y^{2}-2 Y-1=0$ (which are none other than the $2 \cos \frac{2 k \pi}{7}$ for $k$ ranging from 1 to 6, since these six cosines are equal in pairs),