<s>[INST] 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 ?
 [/INST] 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.

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