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?

Hint:
$$\cos3\beta+\sin3\beta=4(\cos^3\beta-\sin^3\beta)-3(\cos\beta-\sin\beta)$$
$$(\cos\beta-\sin\beta)^3=\cos^3\beta-\sin^3\beta-3\cos\beta\sin\beta(\cos\beta-\sin\beta)$$
$$y^2=?$$
Replace the values of $\cos\beta\sin\beta,\cos\beta-\sin\beta$
