Eliminating parameter $\beta$ from $x=\cos 3 \beta + \sin 3 \beta$, $y = \cos \beta - \sin \beta$ 
Based on the given parametric equations:
$$\begin{align}
x &=\cos 3 \beta + \sin 3 \beta \\
y &= \cos \beta \phantom{3}- \sin \beta
\end{align}$$
Eliminate the parameter $\beta$ to prove that $x-3y+2y^3=0$.

What I got so far: 
$$\cos 3 \beta + \sin 3 \beta = ( \cos \beta - \sin \beta)(1+4\sin\beta\cos\beta)$$
Which trigonometric identity should I use to proceed?

Making
$$
y = \frac 12\left(e^{i\beta}+e^{-i \beta}\right)-\frac{1}{2i}\left(e^{i\beta}-e^{-i\beta}\right)
$$
we have after powering and collecting
$$
y^3 = \frac 32\left(\cos\beta-\sin\beta\right)-\frac 12\left(\cos(3\beta)+\sin(3\beta)\right)
$$
then follows
$$
y^3 = \frac 32 y - \frac 12 x
$$
