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?
 A: 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 $\cos\beta\sin\beta,\cos\beta-\sin\beta$
A: 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
$$
y^3 = \frac 32 y - \frac 12 x
$$
A: 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 $\beta$
A: $$y=\sqrt2\cos\left(\dfrac\pi4+\beta\right)=\sqrt2\cos t$$ where $\dfrac\pi4+\beta=t$
$$x= \cos3\left(t-\dfrac\pi4\right)+\sin3\left(t-\dfrac\pi4\right)$$
As $\cos3\dfrac{\pi}4=\cos\left(\pi-\dfrac\pi4\right)=-\cos\dfrac\pi4=?$
$\sin3\dfrac\pi4=?$
$$x=-\dfrac1{\sqrt2}\cos3t+\dfrac1{\sqrt2}\cdot\sin3t-\dfrac1{\sqrt2}\cdot\sin3t-\dfrac1{\sqrt2}\cdot\cos3t$$
$$\implies\sqrt2x=-2\cos3t$$
Use $\cos3t=4\cos^3t-3\cos t$
