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: 
The Law of Cosines is equivalent to (and is often proven via) the statements
$$a = b \cos\gamma + c\cos\beta \qquad b = c \cos\alpha + a \cos\gamma \qquad c = a \cos\beta + b \cos\alpha$$
so, your sum becomes
$$\frac{c \cos\alpha - a \cos\gamma}{a\sin\gamma} + \frac{a \cos\beta - b \cos\alpha}{b\sin\alpha} + \frac{b \cos\gamma - c \cos\beta}{c\sin\beta}$$

Further, the Law of Sines allows us to write
$$a = d \sin\alpha \qquad b = d\sin\beta \qquad c = d \sin\gamma$$
where $d$ is the triangle's circumdiameter. This gives

$$\frac{\sin\gamma \cos\alpha - \sin\alpha \cos\gamma}{\sin\alpha\sin\gamma} + \frac{\sin\alpha \cos\beta - \sin\beta \cos\alpha}{\sin\beta\sin\alpha} + \frac{\sin\beta \cos\gamma - \sin\gamma \cos\beta}{\sin\gamma\sin\beta}$$
$$= \left( \cot\alpha - \cot\gamma \right) + \left( \cot\beta - \cot\alpha \right) + \left( \cot\gamma - \cot\beta \right ) = 0 $$
Consider the first summand, to begin with:
$$
C=\frac{b-2a\cos\gamma}{a\sin\gamma}
$$
The law of cosines tells you that
$$
c^2=a^2+b^2-2ab\cos\gamma
$$
so we can write
$$
b-2a\cos\gamma=\frac{c^2-a^2}{b}
$$
and so we have
$$
C=\frac{c^2-a^2}{ab\sin\gamma}
$$
The law of sines is
$$
\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}=2R
$$
where $R$ is the radius of the circumscribed circle. In particular,
$$
\sin\gamma=\frac{c}{2R}
$$
and so we have
$$
C=\frac{2R(c^2-a^2)}{abc}
$$
Similarly,
$$
A=\frac{c-2b\cos\alpha}{b\sin\alpha}=\frac{a^2-b^2}{abc}\\
B=\frac{a-2c\cos\beta}{c\sin\beta}=\frac{b^2-c^2}{abc}
$$
and
$$
C+A+B=\frac{c^2-a^2+a^2-b^2+b^2-c^2}{abc}=0
$$