Derive $\cos(3\theta)=(4\cos\theta)^3 − 3\cos\theta$ I'm having trouble with the following derivation:
Q: 
We can use Euler's Theorem ($e^{i\theta} = \cos\theta + i\sin\theta$), where $e$ is the base of the natural logarithms, and $i = \sqrt{-1}$, together with the binomial theorem as above, to derive a number of trigonometric identities. E.g., if we consider $(e^{i\theta})^2$, we can evaluate it two different ways. First, we can multiply exponents, obtaining $e^{i \, 2\theta}$ and then applying Euler's formula to get $\cos(2\theta) + i \sin(2\theta)$, or we can apply Euler's formula to the inside, obtaining $(\cos\theta + i \sin\theta)^2$, which we then evaluate via the binomial theorem:
\begin{align}
\cos(2\theta) + i \sin(2\theta) &= (\cos\theta + i \sin\theta)^2 \\
&= (\cos\theta)^2 + 2 i \cos\theta \sin\theta + i^2 (\sin\theta)^2 \\
&=(\cos\theta)^2 + 2 i \cos\theta \sin\theta − (\sin\theta)^2
\end{align}
Equating real and imaginary parts gives us
\begin{align}
\cos(2\theta) &= (\cos\theta)^2 − (\sin\theta)^2 \\ 
\sin(2\theta) &= 2 \cos\theta \sin\theta
\end{align}
We can then rewrite the first of these identities, using $1=(\sin\theta)^2+(\cos\theta)^2$ to get $(\cos\theta)^2=1−(\sin\theta)^2$, whence the familiar
$$\cos(2\theta)=1−2(\sin\theta)^2$$
Use this same approach to show $\cos(3\theta)=(4\cos\theta)^3−3\cos\theta$.
A:
This is my work so far:
\begin{align}
e^{i \, 3\theta} &= \cos(3\theta) + i \sin(3\theta) = (\cos\theta)^3 + 3 i (\cos\theta)^2 \sin\theta - 3\cos\theta (\sin\theta)^2 - i(\sin\theta)^3 \\
\cos(3\theta) &= (\cos\theta)^3 - 3(\sin\theta)^2 \cos\theta \\
\sin(3\theta) &= 3\sin\theta(\cos\theta)^2 - (\sin\theta)^3
\end{align}
But now I'm unsure how to get $\cos(3\theta) = (4 \cos\theta)^3 − 3\cos\theta$ from what I've derived.

I believe your claim is actually incorrect. You should have the identity
\begin{align}
\cos 3\theta = 4\cos^3\theta -3\cos\theta
\end{align}
not
\begin{align}
\cos 3\theta = (4\cos\theta)^3 -3\cos\theta.
\end{align}
Observe
\begin{align}
e^{i3\theta} =&\ (\cos\theta + i\sin\theta)^3 = \cos^3\theta+3(i\sin\theta)\cos^2\theta+3(i\sin\theta)^2\cos\theta +(i\sin\theta)^3\\
=&\ \cos^3\theta -3\sin^2\theta \cos\theta+i(3\sin\theta\cos^2\theta-\sin^2\theta).
\end{align}
Hence taking the real part yields
\begin{align}
\cos 3\theta =&\ \cos^3\theta -3\sin^2\theta \cos \theta\\
=&\ \cos^3\theta - 3(1-\cos^2\theta)\cos \theta\\
=&\ 4\cos^3\theta -3\cos\theta.
\end{align}
