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 ?
 A: 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.
A: 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
$$t_k = 2\sqrt{-p/3}\cos(\phi/3 + 2k\pi/3) , k = 0,1,2 .$$
A: 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:
$$
\begin{aligned}
u\omega +v\bar\omega
&=
r(\cos\theta+ i\sin\theta)(\cos(2\pi/3)+ i\sin(2\pi/3))
\\
&\ +
r(\cos\theta- i\sin\theta)(\cos(2\pi/3)- i\sin(2\pi/3))
\\[2mm]
&=
r(\cos(\theta+2\pi/3) + i\sin(\theta+2\pi/3))
\\
&\ +
r(\cos(-\theta-2\pi/3) + i\sin(-\theta-2\pi/3))
\\[2mm]
&=
2r\cos(\theta+2\pi/3)\ .
\end{aligned}
$$
A: 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,
$$u+v=\sqrt[3]\rho\left(\cos\frac\theta3+i\sin\frac\theta3+\cos\frac\theta3-i\sin\frac\theta3\right)=2\sqrt[3]\rho\cos\frac\theta3$$ for $k=0,1,2$.
A: 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
$$\cos3\theta.$$
