constructibility of heptagon from with straight edge, compass and angle trisector Angle trisector can be described from root of cubic $4x^3-3x=k$ for some constructible number $k$. The root of equation $x^3+x^2-2x-1=0$ are $2\cos (2 \pi/7), 2\cos (4 \pi/7)$ and $2\cos (6 \pi/7)$. How can be describe the construction of regular heptagon algebraically if roots of $4x^3-3x=k$ were constructible? Do those two cubics share root for some choice of constructible $k$? It has already been discussed geometrically, but I don't seem to know the algebraic approach.

This is straightforward

1. Get rid of the quadratic term (=depress the cubic). Let $x=u-1/3$ and we get $$u^3-\frac73u-\frac7{27}=0.$$
2. Scale it correctly to get that $[4:-3]$ ratio for the coefficients of the cubic/linear terms. Let $u=2\sqrt 7v/3$ and arrive at $$\frac{14\sqrt7}{27}\left(4v^3-3v-\frac1{2\sqrt7}\right)=0.$$



As $1/2\sqrt7$ is constructible, the angle trisector gives you $v$, then linear substitutions with constructible coefficients give you $u$ and $x$.