I'm having trouble with part of a question on Cardano's method for solving cubic polynomial equations. This is a multi-part question, and I have been able to answer most of it. But I am having trouble with the last part. I think I'll just post here the part of the question that I'm having trouble with.

We have the depressed cubic equation : \begin{equation} f(t) = t^{3} + pt + q = 0 \end{equation} We also have what I believe is the negative of the discriminant : \begin{equation} D = 27 q^{2} + 4p^{3} \end{equation} We assume $p$ and $q$ are both real and $D < 0$. We also have the following polynomial in two variables ($u$ and $v$) that results from a variable transformation $t = u+v$ : \begin{equation} u^{3} + v^{3} + (3uv + p)(u+v) + q = 0 \end{equation} You also have the quadratic polynomial equation : \begin{equation} x^{2} + qx - \frac{p^{3}}{27} = 0 \end{equation} The solutions to the 2-variable polynomial equation satisfy the following constraints : \begin{equation} u^{3} + v^{3} = -q \end{equation} \begin{equation} uv = -\frac{p}{3} \end{equation} The first section of this part of the larger question asks to prove that the solutions of the quadratic equation are non-real complex conjugates. Here the solutions to the quadratic are equal to $u^{3}$ and $v^{3}$ (this relationship between the quadratic polynomial and the polynomial in two variables was proven in an earlier part of the question). I was able to do this part. The second part of this sub-question is what I'm having trouble with.

The question says, let : \begin{equation} u = r\cos(\theta) + ir\sin(\theta) \end{equation} \begin{equation} v = r\cos(\theta) - ir\sin(\theta) \end{equation} The question then asks the reader to prove that the depressed cubic equation has three real roots : \begin{equation} 2r\cos(\theta) \text{ , } 2r\cos\left( \theta + \frac{2\pi}{3} \right) \text{ , } 2r\cos\left( \theta + \frac{4\pi}{3} \right) \end{equation} In an earlier part of the question they had the reader prove that given : \begin{equation} \omega = \frac{-1 + i\sqrt{3}}{2} \end{equation} s.t. : \begin{equation} \omega^{2} = \frac{-1 - i\sqrt{3}}{2} \end{equation} and : \begin{equation} \omega^{3} = 1 \end{equation} that if $(u,v)$ is a root of the polynomial in two variables then so are : $(u\omega,v\omega^{2})$ and $(u\omega^{2},v\omega)$. I think that the part of the question I'm having trouble with is similar. I suspect that : \begin{equation} 2r \cos\left( \theta + \frac{2\pi}{3} \right) = u\omega + v\omega^{2} \text{ or } u\omega^{2} + v\omega \tag{1} \end{equation} and : \begin{equation} 2r \cos\left( \theta + \frac{4\pi}{3} \right) = u\omega + v\omega^{2} \text{ or } u\omega^{2} + v\omega \tag{2} \end{equation} I have derived that : \begin{equation} \omega = \cos(\phi) + i\sin(\phi) \end{equation} where $\phi = \frac{2\pi}{3}$. Also : \begin{equation} \omega^{2} = \cos(2\phi) + i \sin(2\phi) \end{equation} So that the goal of the question may be to prove equations $(1)$ and $(2)$. I have tried to do this but haven't been able to.

Am I approaching this question in the correct way ? If I am approaching it the right way can someone show me how to use trigonometric identities to prove equations #1 and #2 ?

  • $\begingroup$ You just have to remind that $\omega$ and $\omega^2$ are conjugate. $\endgroup$ – Bernard Jul 29 '20 at 21:25
  • $\begingroup$ Is this really Cardano's technique? It seems to use a fair amount of complex analysis and analytic geometry, which feels much more modern to me (at least post Descartes for the analytic geometry; maybe post Euler or Fourier for the complex analysis?). Do you have a source for this approach? $\endgroup$ – Xander Henderson Jul 29 '20 at 21:27
  • $\begingroup$ @XanderHenderson: As far as I know, what we now call complex numbers were invented by Cardano-Tartaglia precisely to have solutions for the ‘impossible case’ of the method. $\endgroup$ – Bernard Jul 29 '20 at 21:33
  • $\begingroup$ @Bernard Indeed, but my understanding is that the polar form of complex numbers were not understood until much later or used until much later. $\endgroup$ – Xander Henderson Jul 29 '20 at 21:34
  • $\begingroup$ Certainly, but cubic roots of unity were known, I believe (not necessarily under that name). $\endgroup$ – Bernard Jul 29 '20 at 21:36

Suppose that $u$ and $v$ are such that $u^3+v^3=-q$ and that $3uv=-p$. You already know that then $u+v$ is a root of the depressed equation. On the other hand, $u^3$ and $v^3$ are the roots of a quadratic equation with real coefficients and without real roots; it follows that $v^3=\overline{u^3}=\overline u^3$ and that therefore, $v=\overline u$, $v=\omega\overline u$ or $v=\omega^2\overline u$. But, since $3uv=-p\in\Bbb R$, then in fact, you can't have $v=\omega\overline u$ and neither can you have $v=\omega^2\overline u$. Conclusion: $y=\overline u$.

If $u=r(\cos\theta+i\sin\theta)$, then $v=\overline u=r(\cos\theta-i\sin\theta)$, and so $u+v=2\cos\theta$.

Now, let $u'=\omega u$ and let $v'=\omega^2v$. Then $u'^3+v'^3=-q$ and $3u'v'=-p$. So, $u'+v'$ is also a root of the cubic. But\begin{align}u'+v'&=(r\cos\theta+ri\sin\theta)\left(\cos\left(\frac{2\pi}3\right)+\sin\left(\frac{2\pi}3\right)i\right)+\\&\ +(r\cos(-\theta)+ri\sin(-\theta))\left(\cos\left(\frac{-2\pi}3\right)+\sin\left(-\frac{2\pi}3\right)i\right)\\&=2r\cos\left(\theta+\frac{2\pi}3\right).\end{align}

Finally, if you take $u''=\omega^2u$ and $v''=\omega v$, you can deduce that $2r\cos\left(\theta+\frac{4\pi}3\right)$ is still another root of your cubic.


Let $w(\alpha) = \cos \alpha + i\sin \alpha$. Then $$w(\alpha) w(\beta) = (\cos\alpha + i \sin \alpha)(\cos \beta + i\sin \beta) \\ =\cos\alpha \cos \beta - \sin \alpha \sin \beta +i(\cos\alpha \sin \beta + \sin \alpha \cos \beta) = \cos(\alpha + \beta) + i \sin(\alpha + \beta) \\= w(\alpha + \beta) .$$ An easier way to see this is to write $w(\alpha) = e^{i\alpha}$. Then $$w(\alpha) w(\beta) = e^{i\alpha}e^{i\beta} = e^{i(\alpha + \beta)} = w(\alpha + \beta) .$$

We have $$u\omega = rw(\theta)w(\phi) = rw(\theta+\phi) ,$$ $$u\omega^2 = rw(\theta)w(2\phi) = rw(\theta+2\phi) .$$ Moreover, since $v = \overline u$ and $\omega^2 = \overline \omega$, we get $$v\omega^2 = \overline u \cdot \overline \omega = \overline{u\omega} ,$$ thus $$u\omega + v\omega^2 = 2\Re (u\omega) = 2r\cos(\theta + \phi) = 2r\cos(\theta + 2\pi/3) .$$ Similarly $$v\omega = \overline u \cdot \overline {\omega^2} = \overline{u\omega^2},$$ thus $$u\omega^2 + v\omega = 2\Re (u\omega^2) = 2r\cos(\theta + 2\phi) = 2r\cos(\theta + 4\pi/3) .$$


In my opinion it is an odd aproach to apply Cardano's formula and then translate the result into a trigonometric form. A direct approach is via angle trisection. By Moivre's formula we have $$\cos\phi + i\sin\phi = (\cos(\phi/3) + i\sin(\phi/3))^3$$ which gives $$\cos \phi = \cos^3(\phi/3) -3\cos(\phi/3)\sin^2(\phi/3)\\ = \cos^3(\phi/3) -3\cos(\phi/3)(1- \cos^2(\phi/3)) = 4 \cos^3(\phi/3) - 3 \cos(\phi/3) .$$ Writing $\theta = \phi/3$ and $x = 2\cos \theta$ gives us the cubic angle trisection equation $$x^3 - 3x = 2\cos \phi \tag{1}.$$ By construction it has the obvious solution $x_0 = 2\cos \theta$. But since $\cos \phi = \cos (\phi + 2\pi) = \cos (\phi + 4 \pi)$, it also has the solutions $x_1 = 2 \cos((\phi + 2\pi)/3) = 2\cos (\theta + 2\pi/3)$, $x_2 = 2 \cos((\phi + 4\pi)/3) = 2\cos (\theta + 4\pi/3)$.

Under the assumption that $p, q$ are real and $D = 27q^2 + 4 p^3<0$ it is possible to reduce the general equation $$t^3 + pt + q = 0 \tag{2}$$ to the angle trisection equation (1). Since $D < 0$, we must have $p < 0$. Note that therefore $D < 0$ is equivalent to $27q^2/(-4p^3) < 1$.

Let us write $t = cx$. Then $$x^3 + (p/c^2)x = -q/c^3 .$$ With $c = \sqrt{-p/3} > 0$ we get $$x^3 -3x = 2(-q/2c^3) .$$ But $$(-q/2c^3)^2 = q^2 /4(-p/3)^3 = 27q^2/(-4p^3) < 1$$ which means that $$-q/2c^3 \in (-1,1) .$$ Therefore $\phi = \arccos(-q/2c^3)$ is a well-defined number in $(0,2\pi)$ and we get the cubic equation (1) with solutions $x_k$ as above. Therefore the solutions of (2) are $$t_k = 2\sqrt{-p/3}\cos(\phi/3 + 2k\pi/3) , k = 0,1,2 .$$


Here is my way to see the algebra beyond the solution of the cubic. It is based on the known algebraic identity: $$ \tag{$*$} t^3+x^3+y^3-3txy =(t+x+y)(t+\omega x+\omega^2y)(t+\omega^2 x+\omega y)\ . $$ Then with the notations from the OP, taking $x,y$ to be $-u,-v$: $$ \begin{aligned} 0 &=t^3+pt+q\\ &=t^3-3tuv-x^3-y^3\\ &=(t-u-v)(t-\omega u-\omega^2 v)(t-\omega^2 u+\omega v)\ . \end{aligned} $$ So the roots of the cubic are $u\omega^k + v\omega^{2k}=u\omega^k + v\bar\omega^k$, for $k$ among $0,1,2$.

Now consider $u,v$ to be $r(\cos\theta\pm i\sin\theta)$. Then the root $u+v$ is immediately seen to be $2r\cos \theta$.

The other two are equally simple, for instance: $$ \begin{aligned} u\omega +v\bar\omega &= r(\cos\theta+ i\sin\theta)(\cos(2\pi/3)+ i\sin(2\pi/3)) \\ &\ + r(\cos\theta- i\sin\theta)(\cos(2\pi/3)- i\sin(2\pi/3)) \\[2mm] &= r(\cos(\theta+2\pi/3) + i\sin(\theta+2\pi/3)) \\ &\ + r(\cos(-\theta-2\pi/3) + i\sin(-\theta-2\pi/3)) \\[2mm] &= 2r\cos(\theta+2\pi/3)\ . \end{aligned} $$


You can "brute force" the solution of the quadratic,


giving two roots

$$u^3,v^3=\frac{-q\pm\sqrt{q^2+\dfrac{4p^3}{27}}}2$$ which are complex. In polar form,

$$\rho=\frac{q^2}2+\frac{p^3}{27}$$ and $$\theta=\pm\arctan\sqrt{1+\dfrac{4p^3}{27q^2}}+k\pi.$$

Now after taking the cubic roots,

$$u+v=\sqrt[3]\rho\left(\cos\frac\theta3+i\sin\frac\theta3+\cos\frac\theta3-i\sin\frac\theta3\right)=2\sqrt[3]\rho\cos\frac\theta3$$ for $k=0,1,2$.


An alternative method:

We can try to turn $$t^3+pt+q=0$$ into $$4\cos^3\theta-3\cos\theta=a$$ by a change of variable: we set

$$t=\lambda \cos\theta$$ and solve

$$-\frac3{4\lambda^2}=p,$$ or

$$\lambda=\sqrt{-\frac3{4p}}.$$ This establishes


But the LHS is just



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