Let \[ p(z)=a_0z^3+a_1z^2+a_2z+a_3 \] be a cubic polynomial. Assume \(a_0\ne0\) and \(p'(z)\) has distinct zeros.
Show that there are \(A,B\in\operatorname{Aut}(\mathbb C)\) such that \[ A(p(B(z)))=z^3-3z. \]
Show how to solve \[ z^3-3z=w \] by radicals, using the fact that if \(z=a+a^{-1}\), then \[ z^3-3z=a^3+a^{-3}. \]
Combining the previous calculations, solve \[ z^3+az+b=0 \] by radicals.
Solution
1. Let \(r_+\) and \(r_-\) be the two distinct zeros of \(p'\). There is anaffine automorphism \(B\) of \(\mathbb C\) with\[ B(1)=r_+, \qquad B(-1)=r_-. \]
Since \(r_+\neq r_-\), the map \[ B(z)=\frac{r_+-r_-}{2}z+\frac{r_++r_-}{2} \] has nonzero linear coefficient and therefore lies in \(\operatorname{Aut}(\mathbb C)\).
2. If \(q=p\circ B\), then there are constants \(\lambda\neq0\) and \(\mu\)such that\[ q(z)=\frac{\lambda}{3}(z^3-3z)+\mu. \]
By the chain rule, \[ q'(z)=p'(B(z))B'(z). \] Thus the zeros of \(q'\) are exactly \(1\) and \(-1\). Since \(q'\) is a quadratic polynomial with distinct zeros \(\pm1\), it has the form \[ q'(z)=\lambda(z^2-1) \] for some \(\lambda\neq0\). Integrating gives \[ q(z)=\lambda\left(\frac{z^3}{3}-z\right)+\mu =\frac{\lambda}{3}(z^3-3z)+\mu. \]
3. Part (a) follows by taking\[ A(w)=\frac{3}{\lambda}(w-\mu). \]
The map \(A\) is affine with nonzero linear coefficient, hence belongs to \(\operatorname{Aut}(\mathbb C)\). By 2, \[ A(p(B(z))) =\frac{3}{\lambda}\left(q(z)-\mu\right) =z^3-3z. \]
4. To solve\[ z^3-3z=w, \]set \(z=u+u^{-1}\) with \(u\neq0\). Then \(U=u^3\) must satisfy\[ U^2-wU+1=0. \]
The identity supplied in the problem gives \[ (u+u^{-1})^3-3(u+u^{-1})=u^3+u^{-3}. \] Hence the equation becomes \[ u^3+u^{-3}=w. \] Writing \(U=u^3\) and multiplying by \(U\) gives \[ U^2-wU+1=0. \]
5. Therefore the solutions of \(z^3-3z=w\) are obtained by radicals asfollows: choose\[ U=\frac{w+\sqrt{w^2-4}}{2}, \]choose any cube root \(u\) of \(U\), and set\[ \boxed{z=u+u^{-1}}. \]The three choices of the cube root give the three roots, counted withmultiplicity.
The two roots of \(U^2-wU+1=0\) are \[ U_\pm=\frac{w\pm\sqrt{w^2-4}}{2}, \] and satisfy \(U_+U_-=1\), so neither is zero. If \(u^3=U_+\), then \((u^{-1})^3=U_-\). Thus 4 gives \[ (u+u^{-1})^3-3(u+u^{-1})=U_++U_-=w. \] Replacing \(u\) by \(\omega u\), where \(\omega^3=1\), gives all three roots of the cubic, with repetitions precisely when the cubic has a multiple root. Choosing \(U_-\) instead merely replaces \(u\) by a reciprocal and produces the same set of \(z\)-values.
6. For the depressed cubic\[ z^3+az+b=0, \]put\[ \Delta=\frac{b^2}{4}+\frac{a^3}{27}. \]If \(a\neq0\), choose\[ U^3=-\frac b2+\sqrt\Delta, \qquad V=-\frac{a}{3U}. \]Then\[ \boxed{z=U+V} \]is a root.
Because \[ \left(-\frac b2+\sqrt\Delta\right) \left(-\frac b2-\sqrt\Delta\right) =-\frac{a^3}{27}, \] the assumption \(a\neq0\) implies \(U\neq0\), and \[ V^3 =-\frac{a^3}{27U^3} =-\frac b2-\sqrt\Delta. \] Hence \[ UV=-\frac a3, \qquad U^3+V^3=-b. \] Therefore \[ (U+V)^3+a(U+V)+b =U^3+V^3+(3UV+a)(U+V)+b =0. \] This is Cardano’s radical formula, with the cube roots paired by the condition \(UV=-a/3\).
7. All three roots of \(z^3+az+b=0\) are\[ z_k=\omega^kU+\omega^{-k}V, \qquad k=0,1,2, \]where \(\omega=e^{2\pi i/3}\). If \(a=0\), the equation reduces directly to\(z^3=-b\).
For each \(k\), \[ (\omega^kU)(\omega^{-k}V)=UV=-\frac a3 \] and \[ (\omega^kU)^3+(\omega^{-k}V)^3=U^3+V^3=-b. \] The calculation in 6 therefore shows that every \(z_k\) is a root. A cubic has three roots counted with multiplicity, so these give all of them. When \(a=0\), no pairing is needed: the roots are simply the three cube roots of \(-b\).