Equation of three lines forming a equilateral triangle. 
Prove that if general equation $$\cos \left(3\alpha\right)\left(x^3 -3xy^2\right) + \sin3\alpha\left(y^3 - 3x^2y\right) + 3a\left(x^2 + y^2\right) - 4a^3 = 0$$ represents three lines then they form a equilateral triangle of area $3a^2\sqrt 3$.


This equation looks like an good candidate to solve in polar coordinates than in cartesian coordinates.
Converting it into polar coordinates by parts
\begin{align}
\left(x^3 -3xy^2\right) &= r^3 \left(\cos^3 \theta - 3\cos \theta \sin^2 \theta\right) =r^3 \left(4\cos^3 \theta - 3\cos \theta\right) = r^3\cos 3\theta \tag1\label1 \\
\left(y^3 - 3x^2y\right) &= r^3\left(4\sin^3 \theta - 3\sin\theta\right) = r^3\left(\cos\left(3\pi/2 - 3\theta\right)\right) = -r^3\sin 3\theta\tag2\label2
\end{align}
Using $\eqref1$, $\eqref2$
\begin{align}
r^3\left(\cos 3\alpha\cos 3\theta - \sin3\alpha\sin 3\theta\right) + 3ar^2 - 4a^3 &= 0 \\
r^3\cos \left(3\alpha+ 3\theta\right) + 3ar^2 - 4a^3 &= 0
\end{align}
On dividing by $a^3$ and substituting $z = r/a$
$$z^3\cos \left(3\alpha+ 3\theta\right) + 3z^2 - 4 = 0.$$
Now this happens very often to me, I am not able to solve this cubic for $z$.I did try to find the factor by putting $\,z = \sin\left(3\alpha + 3\theta\right),\,\tan\left(3\alpha + 3\theta\right),\,\ldots$  

What should I do now ? how to solve this cubic ?
 A: The given equation always represents three lines. When we rotate the axes by an angle $\alpha$, in polar coordinates, $r$ remains unchanged but $\theta$ changes to $\theta - \alpha$. Thus the combined equation becomes
$$r^3\cos \left(3\alpha+ 3(\theta-\alpha)\right) + 3ar^2 - 4a^3 = 0$$
This is equivalent to rotating the curve clockwise by $\alpha$ about the origin and thus the shape of the curve does not change. Hence it is enough to prove that the above equation represents three lines. 
We have
\begin{align*}
r^3\cos3\theta +3ar^2 - 4a^3 &= 0\\
r^3(4\cos^3 \theta -3\cos\theta) + 3ar^2 - 4a^3 &= 0\\
4x^3 -3x(x^2+y^2)+3a(x^2+y^2) - 4a^3 &= 0\\
x^3 - 3xy^2 + 3ax^2 + 3ay^2 - 4a^3 &= 0
\end{align*}
where in the last two steps, we have converted the equation back to Cartesian. Clearly $x=a$ is a factor and dividing this out we get
$$(x-a)(x^2 - 3y^2 + 4ax + 4a^2) = 0$$
and hence
$$(x-a)((x+2a)^2 - 3y^2) = 0$$
Thus
$$(x-a)(x+2a-\sqrt{3}y)(x+2a +\sqrt{3}y) = 0$$
Now it is easy to see that the circum center of this triangle is the origin and the circum radius is $2a$. Hence the area is $3\sqrt{3}a^2$
A: Real and imaginary parts of $ (x+iy)^3$ can be recognized here.
EDIT1:
Factorization given below tallies fully with the first general equation in OP's post.
$$ (x \cos(\alpha) - y\sin (\alpha) -a)*
(x \cos(\alpha + 2 \pi/3) - y\sin (\alpha + 2 \pi/3) -a) *
(x \cos(\alpha + 4 \pi/3) - y\sin (\alpha + 4 \pi/3) -a)=0\, \tag1 $$
Area = $ \dfrac { \sqrt3 }{4} \cdot { { (2 \sqrt3 a})^2}  = 3 \sqrt3 a^2$
The following is a plot of three  sides for $ \alpha = 12.345, a=1. $ 
BTW, such factorizations are useful in obtaining torsion in non-circular shafts using St. Venant's theory.

A: The solutions are $r = a, \theta = -\alpha$, $r = a, \theta = \frac{2\pi}{3} - \alpha$ and $r = a, \theta = \frac{4\pi}{3} - \alpha$
