Cubic equation solutions form

[Hill]

Homework Statement

Assume that a cubic equation has three real solutions. Show that the solutions of the cubic equation are of the form - see below.

Relevant Equations

Vieta’s formulae, discriminant.

Let's denote $\sqrt[3] r =t$. The three expressions above can be written as $$x_1=2t \cos \frac {\phi} 3, x_2=t (-\cos \frac {\phi} 3 -\sin \frac {\phi} 3), x_3=t (-\cos \frac {\phi} 3 +\sin \frac {\phi} 3)$$ The Vieta's formulae for the given equation are $$x_1+x_2+x_3=0$$ $$x_1 x_2+ x_2 x_3 + x_1 x_3 = p$$ $$x_1 x_2 x_3 = -q$$
The first equation is obviously satisfied by these $x_1, x_2, x_3$. The second becomes $$t^2(2 \cos^2 \frac {\phi} 3 - 1)=p$$ and the third, $$2t^3 \cos \frac {\phi} 3 (2 \cos^2 \frac {\phi} 3 - 1)=-q$$ I hoped to see that the last two equations can be always solved for the $t$ and $\phi$ as required, but I don't.

P.S. I've got a hint and solved it. Closed.

 

[pasmith]

It might help to expand the angles correctly: $$\begin{split} \cos((\phi + 2\pi)/3) = \cos(\phi/3)\cos(2\pi/3) - \sin(\phi/3)\sin(2\pi/3) = - \tfrac12\cos(\phi/3) + \tfrac12 \sqrt{3}\sin(\phi/3) \\ \cos((\phi + 4\pi)/3) = \cos(\phi/3)\cos(4\pi/3) - \sin(\phi/3)\sin(4\pi/3) = - \tfrac12\cos(\phi/3) - \tfrac12 \sqrt{3}\sin(\phi/3) \end{split}$$

 

[pasmith]

Set $\alpha = e^{i\phi/3}$ so that the roots are $$\begin{split} x_1 &= r^{1/3}(\alpha + \bar{\alpha}) \\ x_2 &= r^{1/3}(\alpha\omega + \bar{\alpha}\omega^2) \\ x_3 &= r^{1/3}(\alpha\omega^2 + \bar{\alpha}\omega)\end{split}$$ where $\omega = e^{2\pi i/3}$ satisfies $\omega^2 = \bar{\omega}$ and $1 + \omega + \omega^2 = 0$. Defining $$\begin{split} s_0 &= x_1 + x_2 + x_3 = 0\\ s_1 &= x_1 + \omega x_2 + \omega^2 x_3 = 3r^{1/3} \bar{\alpha} \\ s_2 &= x_1 + \omega^2 x_2 + \omega x_3 = 3r^{1/3}\alpha \end{split}$$ we find that $$\begin{split} s_1^3 + s_2^3 &= -27q = 27r(\alpha^3 + \bar{\alpha}^3) = 54r\cos \phi \\ s_1^3s_2^3 &= (-3p)^3 = 27^2r^2\end{split}$$ and hence $$\begin{split} r^2 &= -\frac{p^3}{27} \\ r\cos\phi &= -\frac{q}{2} \end{split}$$ (Note that $p < 0$ is a necesary condition for $4p^3 + 27q^2 < 0$.)