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}<br /> \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) \\<br /> \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}<br /> x_1 &amp;= r^{1/3}(\alpha + \bar{\alpha}) \\<br /> x_2 &amp;= r^{1/3}(\alpha\omega + \bar{\alpha}\omega^2) \\<br /> x_3 &amp;= 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}<br /> s_0 &amp;= x_1 + x_2 + x_3 = 0\\<br /> s_1 &amp;= x_1 + \omega x_2 + \omega^2 x_3 = 3r^{1/3} \bar{\alpha} \\<br /> s_2 &amp;= x_1 + \omega^2 x_2 + \omega x_3 = 3r^{1/3}\alpha \end{split} we find that <br /> \begin{split}<br /> s_1^3 + s_2^3 &amp;= -27q = 27r(\alpha^3 + \bar{\alpha}^3) = 54r\cos \phi \\<br /> s_1^3s_2^3 &amp;= (-3p)^3 = 27^2r^2\end{split} and hence <br /> \begin{split}<br /> r^2 &amp;= -\frac{p^3}{27} \\<br /> r\cos\phi &amp;= -\frac{q}{2} \end{split} (Note that p &lt; 0 is a necesary condition for 4p^3 + 27q^2 &lt; 0.)