Five Porismatic Equations. Here is a really tough problem.
If
$$\boldsymbol{a\cos\alpha\cos\beta+b\sin\alpha\sin\beta+c=0}$$
$$\boldsymbol{a\cos\gamma\cos\delta+b\sin\gamma\sin\delta+c=0}$$
$$\boldsymbol{a\cos\beta\cos\gamma+b\sin\beta\sin\gamma+c=0}$$
$$\boldsymbol{a\cos\delta\cos\epsilon+b\sin\delta\sin\epsilon+c=0}$$
$$\boldsymbol{a\cos\epsilon \cos\alpha+b\sin\epsilon\sin\alpha+c=0}$$
prove that
$$\boldsymbol{\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=
\left(\frac{1}{b}+\frac{1}{c}\right)
\left(\frac{1}{c}+\frac{1}{a}\right)
\left(\frac{1}{a}+\frac{1}{b}\right)}$$
where all angles are unequal and between $0$ and $2\pi$.
I cannot work out the algebra on this problem.
This is a system of porsimatic equations. Meaning that it only has distinct solutions if some condition on the variables holds.
The method is the following, in the case of a chain of three equations,
$$a\cos\alpha\cos\beta+b\sin\alpha\sin\beta+c=0$$
$$a\cos\beta\cos\gamma+b\sin\beta\sin\gamma+c=0$$
$$a\cos\gamma\cos\alpha+b\sin\gamma\sin\alpha+c=0$$
We can show
$$\tan\frac{1}{2}(\alpha+\beta)=\frac{b}{a}\tan \gamma$$
either by setting an equation in variable $t$ with solutions, $\tan\frac{1}{2}\alpha$ and $\tan\frac{1}{2}\beta$ and using Vietas formulas or more straighforwardly solving for $\sin\gamma$ and $\cos\gamma$ to get $\tan\gamma$.
Similarly
$$\tan\frac{1}{2}(\alpha+\gamma)=\frac{b}{a}\tan \beta$$
$$\tan\frac{1}{2}(\gamma+\beta)=\frac{b}{a}\tan \alpha$$
and using these we get
$$\tan\frac{1}{2}(\alpha-\beta)=\frac{ab\sin(\alpha-\beta)}{a^2\cos\alpha\cos\beta+b^2\sin\alpha\sin\beta}$$ then using the definition of $\tan=\frac{\sin}{\cos}$ and dividing by $\sin\frac{1}{2}(\alpha-\beta)$ we see that
$$a\cos\alpha\cos\beta+b\sin\alpha\sin\beta+c=c-\frac{ab}{a+b}$$
In the case of a chain of five equations we get the fomulas,
$$\tan\frac{1}{2}(\alpha+\gamma)=\frac{b}{a}\tan\beta$$
$$\tan\frac{1}{2}(\beta+\delta)=\frac{b}{a}\tan\gamma$$
$$\tan\frac{1}{2}(\gamma+\epsilon)=\frac{b}{a}\tan\delta$$
$$\tan\frac{1}{2}(\alpha+\delta)=\frac{b}{a}\tan\epsilon$$
$$\tan\frac{1}{2}(\epsilon+\beta)=\frac{b}{a}\tan\alpha$$
But how to proceed from there ?
I guess you want  $\tan\frac{1}{2}(\alpha-\epsilon)
$ as a function of $\alpha$ and $\epsilon$.

(Not a solution. Too long to be a comment. Maybe there are some ideas here to use.)
Let $ \cos \alpha + i \sin \alpha = A $ and similar equations for the rest.
Then, we have $ a ( A + \frac{1}{A} ) ( B + \frac{1}{B} ) - b ( A - \frac{1}{A} ) ( B - \frac{1}{B} ) + 4c = 0 $, and similar.
Clearing denominators gives  $ a( A^2 + 1) ( B^2 + 1 ) - b ( A^2 - 1) ( B^2 - 1) + 4cAB = 0 $, and similar.
For a fixed $A$, view this as a quadratic in $B$:
$$ B^2 [ (a-b)A^2 + (a+b)] + B [ 4Ac ] + [(a+b)A^2 + (a-b) ] = 0 $$
Since $B, E$ are distinct roots of this quadratic, we may apply Vieta's to obtain $ B+E, BE$, and similar equations.
