Worked solution: Wantzel's algebraic impossibility proof via field extensions (1837)

An angle $\theta$ is constructible exactly when the length $\cos\theta$ is constructible, since dropping a perpendicular from a point on the unit circle at angle $\theta$ onto the $x$-axis produces the segment $\cos\theta$, and conversely $\theta$ is recovered from $\cos\theta$ by erecting a perpendicular. Take $\theta = 20^\circ$, so trisecting $60^\circ$ is possible exactly when $20^\circ$, equivalently $\cos 20^\circ$, is constructible.

The triple-angle identity $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ applied at $\theta=20^\circ$ gives $4\cos^3(20^\circ) - 3\cos(20^\circ) = \cos(60^\circ) = \tfrac12$. Multiplying through by $2$ and substituting $x = 2\cos(20^\circ)$ (so $\cos 20^\circ = x/2$) turns this into $8(x/2)^3 - 6(x/2) = 1$, i.e. $x^3 - 3x - 1 = 0$. Since multiplying or dividing by the nonzero rational number $2$ does not change constructibility, $\cos 20^\circ$ is constructible if and only if $x = 2\cos 20^\circ$ is.

So the entire geometric problem has been reduced to a single algebraic question about the integer-coefficient cubic $P(x) = x^3 - 3x - 1$: is its root $x = 2\cos 20^\circ$ constructible? The next step answers this by computing the degree of $P$ over $\mathbb{Q}$.