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 $\epsilon$.

Below is we are ignoring the edge cases such as $\cos\theta = \pm\cos\phi,$ where $\theta,\phi$ are two angles among the given $5.$

$$(a\cos\alpha\cos\beta + c)^2 = b^2\sin^2\alpha\sin^2\beta\implies$$ The quadratic $$f(t) = ((a^2-b^2)\cos^2\beta + b^2)t^2 + 2t\cdot (ac\cos\beta) + (c^2-b^2\sin^2\beta)$$ has roots $t = \cos\alpha$ and $t = \cos\gamma$ and thus: $$\cos\alpha\cos\gamma = \dfrac{c^2-b^2\sin^2\beta}{a^2\cos^2\beta+b^2\sin^2\beta} = \dfrac{c^2\sec^2\beta-b^2\tan^2\beta}{a^2+b^2\tan^2\beta} = \dfrac{c^2 + (c^2-b^2)\tan^2\beta}{a^2+b^2\tan^2\beta}.$$

We use the identity derived in OP's text: $$\tan\frac{1}{2}(\alpha+\gamma) = \dfrac{b}{a}\tan\beta,\quad (1)$$ to obtain: $$\cos\alpha\cos\gamma = \dfrac{c^2+\frac{a^2(c^2-b^2)}{b^2}\left(\tan^2\dfrac{1}{2}(\alpha+\gamma)\right)}{a^2\left(1+\tan^2\dfrac{1}{2}(\alpha+\gamma)\right)}$$ and we can get rid of the square and half-angle above by using: $$\tan^2x = \dfrac{\sin^2x}{\cos^2x} = \dfrac{1-\cos 2x}{1+\cos 2x}.$$ Then we obtain the following: $$\begin{align} \cos\alpha\cos\gamma &= \dfrac{c^2+\dfrac{a^2(c^2-b^2)}{b^2}\cdot\dfrac{1-\cos(\alpha+\gamma)}{1+\cos(\alpha+\gamma)}}{a^2\left(1+\dfrac{1-\cos(\alpha+\gamma)}{1+\cos(\alpha+\gamma)}\right)} \\ &= \dfrac{c^2}{2a^2}(1+\cos(\alpha+\gamma))+\dfrac{c^2-b^2}{2b^2}(1-\cos(\alpha+\gamma))\quad (2) \end{align}$$ Now, we do the spliting $\cos(\alpha+\gamma) = \cos\alpha\cos\gamma - \sin\alpha\sin\gamma$ and plug it into $(2)$, to finally obtain: $$\left(\dfrac{1}{c^2} - \dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\cos\alpha\cos\gamma+\left(\dfrac{1}{c^2} + \dfrac{1}{a^2}-\dfrac{1}{b^2}\right)\sin\alpha\sin\gamma + \left(\dfrac{1}{c^2} - \dfrac{1}{a^2}-\dfrac{1}{b^2}\right) = 0 \quad (3).$$

Now we can finish the problem by noting that the above "shifting" produces following chain of permutations: $$(\alpha,\beta,\gamma, \delta,\epsilon)\longrightarrow (\alpha, \gamma, \epsilon, \beta, \delta) \longrightarrow (\color{red}{\alpha}, \epsilon, \delta, \color{red}{\gamma}, \color{red}{\beta})\longrightarrow\dots$$ This can go on for two more times but we have what we need from the third permutation above because this configuration also gives: $$\dfrac{b_2}{a_2}\tan\beta = \tan\dfrac{\alpha+\gamma}{2} = \dfrac{b}{a}\tan\beta\iff \dfrac{b_2}{a_2} = \dfrac{b}{a},$$ where $b_2,a_2$ are the results of doing the transformation: $$(a,b,c)\longrightarrow \left(\dfrac{1}{c^2}-\dfrac{1}{a^2}+\dfrac{1}{b^2}, \dfrac{1}{c^2}+\dfrac{1}{a^2}-\dfrac{1}{b^2}, \dfrac{1}{c^2}-\dfrac{1}{a^2}-\dfrac{1}{b^2}\right)$$ twice. Here, let's reciprocate $a,b,c\to \frac 1a, \frac 1b, \frac 1c$ to simplify writing. This yields a big equation: $$\begin{align} \dfrac{a}{b} &= \dfrac{(c^2-a^2+b^2)^2(c^2+a^2-b^2)^2 + (c^2-a^2-b^2)^2\left((c^2+a^2-b^2)^2-(c^2-a^2+b^2)^2\right)}{(c^2-a^2+b^2)^2(c^2+a^2-b^2)^2 - (c^2-a^2-b^2)^2\left((c^2+a^2-b^2)^2-(c^2-a^2+b^2)^2\right)}\\ &=\dfrac{\left(c^4-(a^4-b^4)^2\right)^2+4c^2\left(c^2-a^2-b^2\right)^2(a^2-b^2)}{\left(c^4-(a^4-b^4)^2\right)^2-4c^2\left(c^2-a^2-b^2\right)^2(a^2-b^2)} \end{align}$$ Subtract $1$ from both sides and factor out $(a-b)$ to give common denominator: $$\begin{align} \left(c^4-(a^4-b^4)^2\right)^2-4c^2\left(c^2-a^2-b^2\right)^2(a^2-b^2) = 8bc^2\left(c^2-a^2-b^2\right)^2(a+b)\implies \\ \left(c^4-(a^4-b^4)^2\right)^2 =4c^2(c^2-a^2-b^2)^2\left(a^2-b^2+2b(a+b)\right)\implies \\ (c^2-a^2+b^2)^2(c^2+a^2-b^2)^2 = 4c^2(a+b)^2(c^2-a^2-b^2)^2 \quad (4) \end{align}.$$

From here, if $c^2 = (a+b)^2,$ then we can notice that: $$(c^2-a^2+b^2)^2(c^2+a^2-b^2)^2 = 16a^2b^2(a+b)^4 =4c^2(a+b)^2(c^2-a^2-b^2)^2, $$ meaning that we have a factor of $c^2-(a+b)^2$ in (4). Finally then $(4)$ factors as

$$(c^2-a^2+b^2)^2(c^2+a^2-b^2)^2 - 4c^2(a+b)^2(c^2-a^2-b^2)^2 = ((c^2-a^2+b^2)(c^2+a^2-b^2) - 2c(a+b)(c^2-a^2-b^2))\cdot ((c^2-a^2+b^2)(c^2+a^2-b^2) + 2c(a+b)(c^2-a^2-b^2)) = $$ $$ = (c-a-b)(a^3+b^3+c^3 - ab^2-a^2b-bc^2-b^2c - ca^2-c^2a-2abc)\cdot (c+a+b)(-a^3-b^3-c^3 + ab^2+a^2b+bc^2-b^2c - ca^2+c^2a-2abc) =0. $$

What's left now is to rule out the case $c\neq\pm (a+b)$ and then we will have one (or both) of only two options left: $$a^3+b^3+c^3 = (a+b)(b+c)(c+a)$$ or $$a^3+b^3-c^3 = (a+b)(b-c)(a-c).$$

(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.

As Somos has thought, the problem is related to Poncelet's Porism.

It is equivalent that there's a pentagon inscribe in $\Gamma_1:x^2+y^2=1$ and circumscribe on $\Gamma_2: \frac{x^2}{c-b}+\frac{y^2}{c-a}+\frac{1}{a+b}=ux^2+vy^2+w=0$

Generally equation $ax_1x_2+by_1y_2+c=0$ means pointer $(x_1,y_1)$ is at polar of conic section $ax^2+by^2+c=0$ for point $(x_2,y_2)$. When point $(x_1,y_1)$ is moving in one conic section, the polar encloses another conic section (here the $ux^2+vy^2+w=0$.

For example, assume a line L is circumscribe on $\Gamma_2$ at $P_0(x_0,y_0)$ and intersection $\Gamma_1$ at $Q_1(x_1,y_1),Q_2(x_2,y_2)$, we have
$$\begin{cases}ux_0^2+vy_0^2+w=0\\ux_0x_1+vy_0y_1+w=0\\ux_0x_2+vy_0y_2+w=0\\x_1^2+y_1^2=1\\x_2^2+y_2^2=1\end{cases}$$ So we have $y_1=\frac{-w-ux_0x_1}{vy_0}$ and $x_1^2+\left(\frac{-w-ux_0x_1}{vy_0}\right)^2=1$, which means $$(v^2y_0^2+u^2x_0^2)x_1^2+2wux_0x_1+w^2-v^2y_0^2=0$$ Similarly we have $(v^2y_0^2+u^2x_0^2)x_2^2+2wux_0x_2+w^2-v^2y_0^2=0$ so that $x_1,x_2$ are two roots of equation $(v^2y_0^2+u^2x_0^2)X^2+2wux_0X+w^2-v^2y_0^2=0$
By Vieta's formula we have $x_1x_2=\frac{w^2-v^2y_0^2}{v^2y_0^2+u^2x_0^2}$ With similar approach we have $y_1y_2=\frac{w^2-u^2x_0^2}{v^2y_0^2+u^2x_0^2}$.
So $ax_1x_2+by_1y_2+c=\frac{(a+b)w^2+(c-a)v^2y_0^2+(c-b)u^2x_0^2}{v^2y_0^2+u^2x_0^2}=\frac{w+vy_0^2+ux_0^2}{v^2y_0^2+u^2x_0^2}=0$.

According to Poncelet's Porism, we could start from any points in $\Gamma_1$ to form the pentagon ABCDE so that we could choose $\alpha=0$ which is equivalent that A(1,0). so $\cos(\beta)=\cos(\epsilon)=-\frac ca$ which shows that $B(-\frac ca, \frac{\sqrt{a^2-c^2}}a),E(-\frac ca,-\frac{\sqrt{a^2-c^2}}a)$.
By symmetric, we have CD circumscribes on $\Gamma_2$ at $(h,0)=(\pm\sqrt{-\frac{w}{u}},0)$ so $h^2=-\frac wu=\frac{b-c}{a+b}$ so $C(h,\sqrt{1-h^2}), D(h,-\sqrt{1-h^2})$ or $C(h,-\sqrt{1-h^2}), D(h,\sqrt{1-h^2})$ or $\cos(\gamma)=\cos(\delta)=h$
Since $a\cos(\beta)\cos(\gamma)+b\sin(\beta)\sin(\gamma)+c=0$ we have $b^2\sin^2(\beta)\sin^2(\gamma)=(-c-a\cos(\beta)\cos(\gamma))^2$ that's
$$b^2(1-\frac{c^2}{a^2})(1-h^2)=c^2(1-h)^2$$ or $$\frac{1-h}{1+h}=\frac{b^2}{c^2}(1-\frac{c^2}{a^2})$$ or $$\frac{b-c}{a+b}=h^2=\left(\frac{b^2a^2-b^2c^2-a^2c^2}{b^2a^2-b^2c^2+a^2c^2}\right)^2$$

Finally we could simplify the formula into $$(a-b)^2(a+b)c^3+ab(a+b)^2c^2-a^2b^2(a+b)c-a^3b^3=0$$ or we could rewrite it as $$ \frac1{a^3}+\frac1{b^3}-\frac1{c^3}=(\frac1a+\frac1b)(\frac1a-\frac1c)(\frac1b-\frac1c)$$

Now we could choose some data to verify the formula. For example a=2, c=1, b=1.0356563731256918583980883131742380541 satisfy formula above (but not for that of dezdichado ) and we could construct one of the root as below

$$\begin{cases} \alpha=0&\cos(\alpha)=1&\sin(\alpha)=0\\ \beta=2.0943951023931954923084289221863352561&\cos(\beta)=-0.5&\sin(\beta)=-0.47942553860420300027328793521557138809\\ \gamma=-1.4622047455839744317084407839834855485&\cos(\gamma)=0.10837828597562565035637625724555457401&\sin(\gamma)=-0.99410972590000622348950262653193401668\\ \delta=1.4622047455839744317084407839834855480&\cos(\delta)=0.10837828597562565035637625724555457446&\sin(\delta)=0.99410972590000622348950262653193401663\\ \epsilon=-2.0943951023931954923084289221863352561&\cos(\epsilon)=-0.5&\sin(\epsilon)=-0.86602540378443864676372317075293618346 \end{cases}$$

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 $a^{-3}+b^{-3}+c^{-3}=1 \neq 0 = (a^{-1}+b^{-1})(a^{-1}+c^{-1})(b^{-1}+c^{-1})$)

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