Solving a cubic with a given substitution

Question

Struggling to make headway on this problem presented in an old exam for which no solutions are available:

Let $f = t^3 - 3m^2 t + m^3 \in \mathbb{Q}[t]$, where $m$ is a fixed natural number.

The equation $f(x) = 0$ is solved by using the substitution $x = r\cos{\theta}$, (where $r > 0 $) and the identity $\cos{3\theta} \equiv 4\cos^3 {\theta} - 3\cos{\theta} $.

Show that there is a solution when $r = 2m$ and $\cos{3\theta} = -\frac{1}{2}$.

I've tried directly substituting in $x = r\cos{\theta}$ as suggested, however I can't get the equation into a form that I can then solve. I have also approached the solution by using the Vieta substitution, but again I could not get the desired result.

How can I solve this as intended?

Answer 1

You're not actually being asked to "solve" the cubic in the traditional sense of expressing any of its roots in terms of radicals. All you're being asked to do here is to verify that if $\cos3\theta=-{1\over2}$, then $x=2m\cos\theta$ satisfies the equation $f(x)=0$. This is done using the identity $\cos3\theta=4\cos^3\theta-3\cos\theta$, as follows:

$$\begin{align} f(2m\cos\theta) &=(2m\cos\theta)^3-3m^2(2m\cos\theta)+m^3\\ &=8m^3\cos^3\theta-6m^3\cos\theta+m^3\\ &=2m^3\left(4\cos^3\theta-3\cos\theta+{1\over2}\right)\\ &=2m^3\left(\cos3\theta+{1\over2}\right)\\ &=0 \end{align}$$

Answer 2

Note that

$$f(r\cos\theta)=r^3\cos^3\theta-3m^2r\cos\theta+m^3$$

To apply the formula, we need

\begin{align} \frac{r^3}{4}&=\frac{3m^2r}{3}\\ r^2&=4m^2 \end{align}

Take $r=2m$.

$$f(2m\cos\theta)=8m^3\cos^3\theta-6m^3\cos\theta+m^3=m^3(2\cos3\theta+1)$$

and can be solved easily.