How prove this $\alpha+\beta+\gamma=n\pi$ let $\theta\in R$,and $\alpha\neq\beta\neq\gamma$
and such
$$\dfrac{\cos{(\alpha+\theta)}}{\sin^3{\alpha}}=\dfrac{\cos{(\beta+\theta)}}{\sin^3{\beta}}=\dfrac{\cos{(\gamma+\theta)}}{\sin^3{\gamma}}$$
prove 
$$\alpha+\beta+\gamma=n\pi$$
My try: let
$$\dfrac{\cos{(\alpha+\theta)}}{\sin^3{\alpha}}=\dfrac{\cos{(\beta+\theta)}}{\sin^3{\beta}}=\dfrac{\cos{(\gamma+\theta)}}{\sin^3{\gamma}}=k$$
then
$$\cos{(\alpha+\theta)}=k\sin^3{\alpha},\quad
\cos{(\beta+\theta)}=k\sin^3{\beta},\quad\cos{(\gamma+\theta)}=k\sin^3{\gamma}$$
so
$$\cos{(\alpha-\beta)}=\cos{[(\alpha+\theta)-(\beta+\theta)]}=\cos{(\alpha+\theta)}\cos{(\beta+\theta)}+\sin{(\alpha+\theta)}\sin{(\beta+\theta)}$$
and follow maybe can't work.Thank you 
 A: Let the ratios are $=k$
$$\implies\cos(\alpha+\theta)=k\sin^3\alpha$$
$$\implies\cos\alpha\cos\theta-\sin\alpha\sin\theta=k\sin^3\alpha$$
Dividing  either sides by $\cos\alpha,$
$$\cos\theta-\tan\alpha\sin\theta=k\tan\alpha\sin^2\alpha=\frac{k\tan\alpha}{\csc^2\alpha}=\frac{k\tan\alpha}{1+\cot^2\alpha}$$
$$\implies \cos\theta-\tan\alpha\sin\theta=\frac{k\tan^3\alpha}{1+\tan^2\alpha} $$
Rearrange to form a cubic equation in $\tan\alpha,$
$$\displaystyle(k+\sin\theta)\tan^3\alpha-\cos\theta\tan^2\alpha+\sin\theta \tan\alpha-\cos\theta=0$$
Observe that $\tan\beta,\tan\gamma$ also satisfy the cubic equation
$\displaystyle\implies \tan\alpha,\tan\beta,\tan\gamma$ are the roots of  $$\displaystyle(k+\sin\theta)t^3-t^2\cos\theta+t\sin\theta-\cos\theta=0$$
Using Vieta's formulas, 
$\displaystyle\tan\alpha+\tan\beta+\tan\gamma=\frac{\cos\theta}{k+\sin\theta}$
and 
$\displaystyle\tan\alpha\tan\beta\tan\gamma=\frac{\cos\theta}{k+\sin\theta}$
$\displaystyle\implies\sum\tan\alpha=\prod\tan\alpha$
Now we know this
A: The question as written allows the solution $\beta = \alpha + m\pi$ and $\gamma = \alpha + n\pi$ for non-zero integers $m\neq n$ and any $\alpha \neq k\pi$. In the following, I'll rule-out this solution.

Write
$$a := e^{2i\alpha} \qquad b := e^{2i\beta} \qquad c := e^{2i\gamma} \qquad t := e^{2i\theta}$$
Presumably, the sines of the angles $\alpha$, $\beta$, $\gamma$ are non-zero, so we have $a \neq 1$ (likewise for $b$ and $c$). We'll also assume $a$, $b$, $c$ to be distinct (in order to avoid the solution in the preamble).
Then invoking the formulas $\cos x = \frac{1}{2}\left( e^{ix} + e^{-ix}\right)$ and $\sin x = \frac{1}{2i}\left( e^{ix} - e^{-ix} \right)$, the initial equation becomes (after some cancellation)
$$\frac{a \left( a t + 1 \right)}{\left( a - 1 \right)^3} 
= \frac{b \left( b t + 1 \right)}{\left( b - 1 \right)^3} 
= \frac{c \left( c t + 1 \right)}{\left( c - 1 \right)^3}$$
The first equality (with $a\neq b$) and second equality (with $b\neq c$) imply,
$$\frac{- 1 + 3 a b - a^2 b - a b^2}{a + b - 3 a b + a^2 b^2} = t = \frac{- 1 + 3 b c - b^2 c - b c^2}{b + c - 3 b c + b^2 c^2}$$
Ignoring $t$ (and assuming $c\neq a$), this gives
$$1 = a b c = e^{2i(\alpha+\beta+\gamma)}$$
so that
$$\alpha + \beta + \gamma = n \pi$$
