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 $\epsilon$.
Best Answer
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 $$a^3+b^3-c^3 = (a+b)(b-c)(a-c).$$