Question: I am preparing for faculty entrance exam and this was the question for which I couldn't find the way to solve (answer is 0). I guess they ask me to solve this by using the rule of sine and cosine: 

Let $\alpha$, $\beta$ and $\gamma$ be the angles of arbitrary triangle with sides a, b and c respectively. Then $${b - 2a\cos\gamma \over a\sin\gamma} + {c-2b\cos\alpha \over b\sin\alpha} + {a - 2c\cos\beta \over c\sin\beta}$$ is equal to (answer is zero but I need steps).

Answer: 
$\displaystyle \sum\_{cyclic} \dfrac{b-2a\cos \gamma}{a\sin \gamma}=\displaystyle \sum\_{cyclic} \dfrac{b^2-2ab\cos \gamma}{ab\sin \gamma}= \dfrac{1}{2S}\displaystyle \sum\_{cyclic} (b^2-2ab\cos \gamma)=0$ by the Cosine Law.
The law of cosines and sines are to be used, your guess is right.

At first consolidate the cosines by summing up with circular symmetry

$$ c^2=a^2+b^2-2ab\cos\gamma + two more $$ gives

$$ a^2 + b^2 + c^2 = 2 ( b\, c \cos\gamma + ... + ....) \tag {1} $$

For sines consolidation it needs multiplying first term numerator and denominator by D where $ c/ \sin \gamma = D, $ the circumcircle diameter.

The first given term becomes

$$ \dfrac{D}{a\,c}(...) $$

All numerators and all denominators can be added separately and placed as numerator and denminator of an equivalent fraction like in:

$$ \frac pq = \frac rs = \frac tu = \frac {p+r+t}{q+s+u} $$

Combine this result with (1) to simplify it to zero.