Problem source

Prove that $\cos3\theta=4\cos^{3}\theta-3\cos\theta$. 
Show how the cubic equation 
\[
24x^{3}-72x^{2}+66x-19=0\tag{*}
\]
can be reduced to the form 
\[
4z^{3}-3z=k
\]
by means of the substitution $y=x+a$ and $z=by$, for suitable values
of the constants $a$ and $b$. Hence find the three roots of the
equation $(*)$, to three significant figures.
Show, by means of a counterexample, or otherwise, that not all cubic
equations of the form 
\[
x^{3}+\alpha x^{2}+\beta x+\gamma=0
\]
can be solved by this method.

Solution source

\begin{align*}
\cos 3\theta &= \cos 2\theta\cos\theta - \sin 2\theta \sin \theta \\
&= (2\cos^2\theta-1)\cos \theta - 2\cos \theta \sin^2 \theta \\
&= 2\cos^3\theta-\cos \theta  - 2\cos \theta(1- \cos^2 \theta) \\
&= 4\cos^3 \theta - 3\cos \theta
\end{align*}

\begin{align*}
0 &= 24x^{3}-72x^{2}+66x-19 \\
&= 24(y+1)^3-72(y+1)^2+66(y+1)-19 \\
&= 24(y^3+3y^2+3y+1)-72(y^2+2y+1)+66(y+1)-19\\
&= 24y^3+(72-144+66)y+(24-72+66-19) \\
&= 24y^3-6y-1 \\
&= 24b'^3z^3 - 6b'z - 1 \\
&= \frac{2}{\sqrt{3}}(4 z^3 -3z) - 1 \\ 
\end{align*}

Therefore if $b = \sqrt{3}, a = 1$, we have:

$4z^3 - 3z = \frac{\sqrt{3}}{2}$

So if $z = \cos \theta \Rightarrow \cos 3\theta = \frac{\sqrt{3}}2 \Rightarrow 3 \theta = \frac{\pi}{6}, \frac{11\pi}{6}, \frac{13\pi}{6} \Rightarrow \theta = \frac{\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}$.

Since $\frac{\cos x}{\sqrt{3}} < 1$ we only need to approximate the first part to 2 significant figures. Therefore:

\begin{align*}
\sqrt{3} &\approx  1 + \frac{1}{1 + \frac{1}{2+\frac11}} = \frac{7}{4} = 1.75\\
\cos \tfrac{\pi}{18} &\approx \cos \frac{1}{6} \approx 1 - \frac{1}{2} \frac{1}{6^2} = \frac{71}{72} \approx 1 - \frac{1}{70} = 1 - 0.014 = 0.986 \\
\frac{\cos \tfrac{\pi}{18}}{\sqrt{3}} & \approx \frac{.986}{1.75} = 0.57 \\
\end{align*}

Final answers: $1.57, 0.803, 0.629$.

We wouldn't be able to solve $x^3 + 1= 0$ using this method, as we would have 2 non-real roots