# Question on Cardano's Method of Solving Cubic Polynomial Equations

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 ?

• You just have to remind that $\omega$ and $\omega^2$ are conjugate. Jul 29, 2020 at 21:25
• 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? Jul 29, 2020 at 21:27
• @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. Jul 29, 2020 at 21:33
• @Bernard Indeed, but my understanding is that the polar form of complex numbers were not understood until much later or used until much later. Jul 29, 2020 at 21:34
• Certainly, but cubic roots of unity were known, I believe (not necessarily under that name). Jul 29, 2020 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 $$u\omega^2 + v\omega = 2\Re (u\omega^2) = 2r\cos(\theta + 2\phi) = 2r\cos(\theta + 4\pi/3) .$$

Edited:

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,

$$x^2+qx-\frac{p^3}{27}=0,$$

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,

$$u+v=\sqrt\rho\left(\cos\frac\theta3+i\sin\frac\theta3+\cos\frac\theta3-i\sin\frac\theta3\right)=2\sqrt\rho\cos\frac\theta3$$ 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

$$4\cos^3\theta-3\cos\theta=-4q\lambda^3.$$

But the LHS is just

$$\cos3\theta.$$