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: Write your system in the form
$$x=\sqrt{2}\sin\left(\frac{\pi}{4}-\beta\right)(1+2\sin(2\beta))$$
$$y=\sqrt{2}\sin\left(\frac{\pi}{4}-\beta\right)$$
then you will get $$x=y(1+2\sin(2\beta))$$ and you can eliminate $\beta$
