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 $\epsilon$.
 A: (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.
A: This is only a small correction to the question statement, but is still too long for the usual comment format. You probably forgot to include the condition that no two of the angles are opposites. Otherwise, there are several simple enough counterexamples, such as the following one.
For simplicity, I write $\theta_1,\ldots,\theta_5$ instead of $\alpha,\beta,\gamma,\delta,\epsilon$, and I also put $c_k=\cos(\theta_k),s_k=\sin(\theta_k)$. The counterexample has $a=-1,b=c=1$ (so $a^{-3}+b^{-3}+c^{-3}=1 \neq 0 = (a^{-1}+b^{-1})(a^{-1}+c^{-1})(b^{-1}+c^{-1})$)
$$
\begin{array}{|c|c|c|c|c|}
\hline
k & c_k & s_k & c_kc_{k+1} & s_ks_{k+1} \\
\hline
1 & -\frac{35}{37} & -\frac{12}{37} & \frac{1925}{2701} & -\frac{576}{2701} \\
\hline
2 & -\frac{55}{73} & \frac{48}{73} & -\frac{275}{949} & \frac{576}{949} \\
\hline
3 & \frac{5}{13} & \frac{12}{13} & -\frac{3}{13} & -\frac{48}{65} \\
\hline
4 & -\frac{3}{5} & -\frac{4}{5} & \frac{21}{37} & -\frac{48}{185} \\
\hline
5 & -\frac{35}{37} & \frac{12}{37} & \frac{1225}{1369} & -\frac{144}{1369} \\
\hline
\end{array}
$$
A: I had originally thought that this question is very similar to other
Porisms such as
Poncelet's closure theorem. After some more work, I think that
it is more similar to non-linear recursions such as the generalized
Lyness recursion. That is $\,x_n = (a+x_{n-1})/x_{n-2}\,$ where $\,a\,$
is a fixed constant. If $\,a=1\,$ the sequence is 5-periodic no matter
what the initial values $\,a_0,a_1\,$ are. If $\,a=0\,$ the sequence is
6-periodic. Otherwise, the period (if any) depends on the initial values.
I think that something similar holds for the system of equations in the
question.
This answer is not yet a complete solution but is a promising approach.
As motivation, consider the trigonometric identities
$$ 4\, t(x)t(y)t(z) = t(-x-y-z) + t(-x+y+z) + t(+x-y+z) + t(+x+y-z) $$
which holds for both $\,t = \sin\,$ and $\,t = \cos.\,$
Thus, define the trigonometric functions:
$$ f(x) := \sin(x+\pi/4) = (\sin(x)+\cos(x))/\sqrt{2} $$
and
$$ F(x,y,z) := (f(-x-y-z) + f(-x+y+z) + f(+x-y+z) + f(+x+y-z))/\sqrt{8}. $$
Notice that $\,F\,$ is fully symmetric and, in general, every equation
$\, F(x,y,z) = t \,$ has two solutions for each of $\,x,y,z\,$
given the other three variables.
The question about the system of equation can be reformulated.
Consider two sequences of real numbers $\,x_n\,$ and $\,w_n\,$ such that the fundamental equation
$$ c_k := F(x_n,x_{n+k},w_k) $$
depends only on $\,k.\,$ Notice that the $\,\{x_0,x_1,\dots\}\,$
correspond to the $\,\{\alpha,\beta,\gamma,\dots\}\,$ of the
question  while $\,(\cos(w_1),\sin(w_1),-c_1)\,$ corresponds to $\,(a,b,c).\,$
Such sequences can be constructed by recursion. Start with
values of $\,x_0,x_1,w_1\,$ and where
$\, c_1 =F(x_0,x_1,w_1).\,$ With the
given values of $\,x_1,w_1,c_1\,$ there are two solutions of
$\, c_1 = F(x,x_1,w_1).$ One solution is $\,x_0\,$ and define $\,x_2\,$
to be the other solution. This provides a recursion for the sequence
$\,x_n\,$ where given $\,x_{n-2}\,$ and $\,x_{n-1}\,$ the $\,x_n\,$ is
the solution of an equation.
Now the problem is to show that there exists $\,c_2\,$ and $\,w_2\,$
such that the fundamental equation holds. So far, I have a computer
algebra proof of this result. Even more challenging is to prove the
fundamental equation for general $\,k>2.\,$
Note that $\,c_0 = 1/\sqrt{2}, w_0 = \pi/4.\,$ Suppose that $\,w_1\,$ is such
that $\,w_N = w_0\,$ for some positive $\,N.\,$ Then the $\,x_n\,$
sequence is periodic with period $\,N\,$ but this depends on the initial
values $\,x_0,x_1.\,$
The question posed is the special where $\,N=5.\,$
