Proving trig identity using De Moivre's Theorem 

Question: Prove $$\cos(3x) = \cos^3(x) - 3\cos(x)\sin^2(x) $$ by using De'Moivres Theorem


So far (learning complex numbers at the moment) that De Moivre's theorem states that
if $z$ $=$ $r\text{cis}(\theta)$ then $z^n = r^n\text{cis}(n\theta)$
so with this question I was thinking if
$$ z = \cos(3\theta) + i\sin(3\theta) $$
then 
$$ z = (\cos(\theta) + i\sin(\theta))^3 $$
and then expanding and comparing the real part? Is that the right way to go for this question?

From Euler's Formula, we have
$$e^{i\theta}=\cos(\theta)+i\sin(\theta) \tag 1$$
Letting $\theta \to n\theta$ in $(1)$ reveals that
$$e^{in\theta}=\cos(n\theta)+i\sin(n\theta) \tag 2$$
Since $e^{in\theta}=\left(e^{i\theta} \right)^n$, then we have from $(1)$ and $(2)$
$$\left(\cos(\theta)+i\sin(\theta)\right)^n=\cos(n\theta)+i\sin(n\theta) \tag 3$$
which is De Moivre's Fomula.  
Finally, letting $n=3$ in $(3)$ and taking the real part reveals 
$$\cos(3\theta)=\text{Re}\left(\cos(\theta)+i\sin(\theta)\right)^3=\cos^3(\theta)-3\cos(\theta)\sin^2(\theta)$$
And we are done!
