Use the identity $\cos 3\theta = 4 \cos^3\theta- 3 \cos \theta$ to solve the cubic equation $t^3 + pt + q = 0$ when $p, q \in \mathbb{R}$. I'm self studying Ian Stewart's Galois Theory and this is Exercise 1.8 from his Third Edition:

Use the identity $\cos 3\theta  = 4 \cos^3\theta- 3 \cos \theta$ to
  solve the cubic equation $t^3 + pt + q = 0$ when $p, q \in \mathbb{R}$
  such that $27 q^2 + 4p^3 < 0$.

I read through many times his method of solving the cubic equation where he didn't use the identity above; yet I'm not sure where the identity can come into play. 
His method is sketched below:
First, he substitutes $t = \sqrt[3]{u} + \sqrt[3]{v}$ and express $t^3$ in terms of $u$ and $v$ as well. Then plugging $t$ and $t^3$ in terms of $u$ and $v$ back to the original equation. Finally solving for $u$ and $v$ will immediately give the zeros. 
Thanks very much for hints and help! 

Note that, since $27q^2+4p^3<0$ and since $27q^2\geqslant0$, $p<0$. So, it makes sense to define $u=\sqrt{-\frac43p}$. Consider the substitution $t=u\cos\theta$. Then $t^3+pt+q$ becomes $u^3\cos^3\theta+pu\cos\theta+q$ and\begin{align}u^3\cos^3\theta+pu\cos\theta+q=0&\iff\frac{u^3\cos^3\theta+pu\cos\theta+q}{u^3/4}=0\\&\iff4\cos^3\theta+\frac{4p}{u^2}\cos\theta+\frac{4q}{u^3}=0\\&\iff4\cos^3\theta-3\cos\theta=-\frac{4q}{u^3}.\end{align}But$$\left(-\frac{4q}{u^3}\right)^2=\frac{16q^2}{(u^2)^3}=\frac{16q^2}{-\frac{64}{27}p^3}=\frac{27q^2}{-4p^3}$$and\begin{align}27q^2+4p^3<0&\iff27q^2<-4p^3\\&\iff\frac{27q^2}{-4p^3}<1\end{align}and therefore $-\frac{4q}{u^3}\in(-1,1)$. So, there is some $\theta\in\Bbb R$ such that $\cos(3\theta)=-\frac{4q}{u^3}$; just take $\theta=\frac13\arccos\left(-\frac{4q}{u^3}\right)$. Then, since $4\cos^3\theta-3\cos\theta=\cos(3\theta)$, $\cos\theta$ is a root of the cubic $t^3+pt+q=0$. But you also have$$\cos\left(3\left(\theta+\frac{2\pi}3\right)\right)=-\frac{4q}{u^3}\quad\text{and}\quad\cos\left(3\left(\theta+\frac{4\pi}3\right)\right)=-\frac{4q}{u^3},$$and therefore $\cos\left(\theta+\frac{2\pi}3\right)$ and $\cos\left(\theta+\frac{4\pi}3\right)$ are also roots of that cubic.
