Prove $\cos 3x =4\cos^3x-3\cos x$

$$e^{ix}=\cos x+i\sin x\Rightarrow e^{3ix}=(\cos x+i\sin x)^3\Rightarrow\cos 3x+i\sin 3x=(\cos x+i\sin x)^3$$

Now expand the cube and equate the real and imaginary parts of both sides to get the answer.

\begin{eqnarray} \cos(3x) &=& \cos(2x+x)\\ &=& \cos(2x)\cos x - \sin(2x)\sin x\\ &=& (\cos^2 x - \sin^2 x)\cos x - 2\sin^2 x\cos x\\ &=& \cos^3 x -(1-\cos^2 x)\cos x - 2 (1 -\cos^2 x)\cos x\\ &=& 2 \cos^3 x -\cos x + \cos^3 x -2 \cos x\\ &=& 4 \cos^3 x - 3 \cos x \end{eqnarray}

I might as well: there is in fact a useful recurrence (Chebyshev) that you can use for expressing $\cos\,nx$ entirely in terms of powers of $\cos\,x$:

$$\cos((n+1)x)=2\cos(x)\cos(nx)-\cos((n-1)x)$$

In this case, you can start with $\cos\,x$ and $\cos\,2x=2\cos^2 x-1$:

$$\begin{align*} \cos\,3x&=(2\cos\,x)(2\cos^2 x-1)-\cos\,x\\ &=4\cos^3 x-2\cos\,x-\cos\,x=4\cos^3 x-3\cos\,x \end{align*}$$