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 $\epsilon$.

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 $a^{-3}+b^{-3}+c^{-3}=1 \neq 0 = (a^{-1}+b^{-1})(a^{-1}+c^{-1})(b^{-1}+c^{-1})$)
$$
\begin{array}{|c|c|c|c|c|}
\hline
k & c_k & s_k & c_kc_{k+1} & s_ks_{k+1} \\
\hline
1 & -\frac{35}{37} & -\frac{12}{37} & \frac{1925}{2701} & -\frac{576}{2701} \\
\hline
2 & -\frac{55}{73} & \frac{48}{73} & -\frac{275}{949} & \frac{576}{949} \\
\hline
3 & \frac{5}{13} & \frac{12}{13} & -\frac{3}{13} & -\frac{48}{65} \\
\hline
4 & -\frac{3}{5} & -\frac{4}{5} & \frac{21}{37} & -\frac{48}{185} \\
\hline
5 & -\frac{35}{37} & \frac{12}{37} & \frac{1225}{1369} & -\frac{144}{1369} \\
\hline
\end{array}
$$
