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).
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
$$\frac{\sin\gamma \cos\alpha - \sin\alpha \cos\gamma}{\sin\alpha\sin\gamma} + \frac{\sin\alpha \cos\beta - \sin\beta \cos\alpha}{\sin\beta\sin\alpha} + \frac{\sin\beta \cos\gamma - \sin\gamma \cos\beta}{\sin\gamma\sin\beta}$$ $$= \left( \cot\alpha - \cot\gamma \right) + \left( \cot\beta - \cot\alpha \right) + \left( \cot\gamma - \cot\beta \right ) = 0 $$
$\displaystyle \sum_{cyclic} \dfrac{b-2a\cos \gamma}{a\sin \gamma}=\displaystyle \sum_{cyclic} \dfrac{b^2-2ab\cos \gamma}{ab\sin \gamma}= \dfrac{1}{2S}\displaystyle \sum_{cyclic} (b^2-2ab\cos \gamma)=0$ by the Cosine Law.
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 $$ C+A+B=\frac{c^2-a^2+a^2-b^2+b^2-c^2}{abc}=0 $$
$\dfrac{b - 2a\cos{\gamma} }{a\sin{\gamma}}=\dfrac{\sin{\beta}}{\sin{\alpha}\sin{\gamma}}-\dfrac{2\cos{\gamma}}{\sin{\gamma}}$
$\sum\dfrac{b - 2a\cos{\gamma} }{a\sin{\gamma}}=\sum \dfrac{\sin^2{\alpha}}{\sin{\alpha}\sin{\beta}\sin{\gamma}}-\sum \dfrac{2\cos{\gamma}\sin{\alpha}\sin{\beta}}{\sin{\alpha}\sin{\beta}\sin{\gamma}}= \dfrac{\sum\sin{\alpha}(\sin{\alpha}-2\cos{\gamma}\sin{\beta})}{\sin{\alpha}\sin{\beta}\sin{\gamma}}$
$\sin{\alpha}(\sin{\alpha}-2\cos{\gamma}\sin{\beta})=\sin{(\beta+\gamma)}(\sin{(\beta+\gamma)}-2\cos{\gamma}\sin{\beta})=\sin{(\beta+\gamma)}(\cos{\beta}\sin{\gamma}-\cos{\gamma}\sin{\beta})=\sin{(\beta+\gamma)}\sin{(\gamma-\beta)}=\dfrac{1}{2}(\cos{2\beta}-\cos{2\gamma}) \implies \sum\sin{\alpha}(\sin{\alpha}-2\cos{\gamma}\sin{\beta})=\dfrac{1}{2}\sum (\cos{2\beta}-\cos{2\gamma}) =0$
From Sine Rule in the triangle, we have $$\frac{\sin\alpha}{a}=\frac{\sin\beta}{b}=\frac{\sin\gamma}{c}=k \space (\text{let any arbitrary constant})$$ Thus by substituting the values of $a=k\sin \alpha$, $b=k\sin \beta$ & $c=k\sin \gamma$ in the given expression, we get $$\frac{b-2a\cos\gamma}{a\sin\gamma}+\frac{c-2b\cos\alpha}{b\sin\alpha}+\frac{a-2c\cos\beta}{c\sin\beta}$$ $$=\frac{k\sin\beta-2k\sin\alpha\cos\gamma}{k\sin\alpha\sin\gamma}+\frac{k\sin\gamma-2k\sin\beta\cos\alpha}{k\sin\beta\sin\alpha}+\frac{k\sin\alpha-2k\sin\gamma\cos\beta}{k\sin\gamma\sin\beta}$$ $$=\frac{\sin\beta-2\sin\alpha\cos\gamma}{\sin\alpha\sin\gamma}+\frac{\sin\gamma-2\sin\beta\cos\alpha}{\sin\beta\sin\alpha}+\frac{\sin\alpha-2\sin\gamma\cos\beta}{\sin\gamma\sin\beta}$$ $$=\frac{\sin(\pi-(\alpha+\gamma))-2\sin\alpha\cos\gamma}{\sin\alpha\sin\gamma}+\frac{\sin(\pi-(\alpha+\beta))-2\sin\beta\cos\alpha}{\sin\beta\sin\alpha}+\frac{\sin(\pi-(\beta+\gamma))-2\sin\gamma\cos\beta}{\sin\gamma\sin\beta}$$ $$=\frac{\sin\alpha\cos\gamma+\cos\alpha\sin\gamma-2\sin\alpha\cos\gamma}{\sin\alpha\sin\gamma}+\frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta-2\sin\beta\cos\alpha}{\sin\beta\sin\alpha}+\frac{\sin \beta\cos\gamma+\cos\beta\sin\gamma-2\sin\gamma\cos\beta}{\sin\gamma\sin\beta}$$ $$=\frac{\sin\gamma\cos\alpha-\cos\gamma\sin\alpha}{\sin\alpha\sin\gamma}+\frac{\sin\alpha\cos\beta-\cos\alpha\sin\beta}{\sin\beta\sin\alpha}+\frac{\sin \beta\cos\gamma-\cos\beta\sin\gamma}{\sin\gamma\sin\beta}$$
$$=(\cot\alpha-\cot\gamma)+(\cot\beta-\cot\alpha)+(\cot\gamma-\cot\beta)$$ $$=0$$
First lets consider $$\frac{b-2a\cos\gamma}{a\sin\gamma}$$ The numerator looks similar to the RHS of the cosine law, but not quite. It would be nice to see $-2ab\cos\gamma$ instead of $-2a\cos\gamma$. So lets just multiply by $b$. Then the numerator would look like this $$b^2-2ab\cos\gamma$$ Now all that is missing is the $a^2$, so just add it $$a^2+b^2-2ab\cos\gamma$$ Now my expression is equal to $c^2$. Easy, right?
It would be nice if math was really like this, but unfortunately, we can't just add and multiply arbitrary constants whenever it seems convenient to do so. We can, however, add zero (e.g. $+\phi-\phi$) and multiply by one (e.g. $\frac{\phi}{\phi}$).
$$\frac{b-2a\cos\gamma}{a\sin\gamma} = \frac{b-2a\cos\gamma}{a\sin\gamma}\cdot\frac{b}{b}=\frac{b^2-2ab\cos\gamma}{ab\sin\gamma}=\frac{a^2+b^2-2ab\cos\gamma-a^2}{ab\sin\gamma}=\frac{c^2-a^2}{ab\sin\gamma}$$ We have just added zero ($a^2-a^2$), so next, lets multiply by one ($\frac{c}{c}$), so we can isolate $\frac{c}{\sin\gamma}$. $$\frac{c^2-a^2}{ab\sin\gamma}=\frac{c^2-a^2}{ab\sin\gamma}\cdot\frac{c}{c}=\frac{c^2-a^2}{abc}\cdot\frac{c}{\sin\gamma}$$ Keeping in mind, of course that $$\Psi=\frac{c}{\sin\gamma}=\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}$$ Then our expression reduces to $$\frac{b-2a\cos\gamma}{a\sin\gamma}+\frac{c-2b\cos\alpha}{b\sin\alpha}+\frac{a-2c\cos\beta}{c\sin\beta}=\frac{c^2-a^2}{abc}\cdot\frac{c}{\sin\gamma}+\frac{a^2-b^2}{abc}\cdot\frac{a}{\sin\alpha}+\frac{b^2-c^2}{abc}\cdot\frac{b}{\sin\beta}$$ $$=\frac{\Psi}{abc}\big((c^2-a^2)+(a^2-b^2)+(b^2-c^2)\big)=0$$
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.
Step 1:
Write cosines relations for a triangle:
$$a^2=b^2+c^2-2bc \cos(A)$$ $$b^2=a^2+c^2-2ac \cos(B)$$ $$c^2=a^2+b^2-2ab \cos(C)$$
Step 2:
Write sinus Area relation for the triangle:
$$Area(ABC)=\frac{ab \sin(C)}{2}=\frac{bc \sin(A)}{2}=\frac{ac \sin(B)}{2}$$
Step 3:
Write cosines relations as you have in your question:
$$a^2-b^2=c(c-2b \cos(A))$$ $$c-2b \cos(A)=\frac{a^2-b^2}{c}$$
$$b^2-c^2=a(a-2c \cos(B))$$ $$a-2c \cos(B)=\frac{b^2-c^2}{a}$$
$$c^2-a^2=b(b-2a \cos(C))$$ $$b-2a \cos(C)=\frac{c^2-a^2}{b}$$
Step 4:
Combine with sinus Area relation
$$c-2b \cos(A)=\frac{a^2-b^2}{c}$$ $$\frac{c-2b \cos(A)}{b \sin(A) }=\frac{a^2-b^2}{c b \sin(A) }$$
$$a-2c \cos(B)=\frac{b^2-c^2}{a}$$ $$\frac{a-2c \cos(B)}{c \sin(B) }=\frac{b^2-c^2}{ac \sin(B)}$$
$$b-2a \cos(C)=\frac{c^2-a^2}{b}$$ $$\frac{b-2a \cos(C)}{a \sin(C) }=\frac{c^2-a^2}{ab \sin(C)}$$
Step 5:
Combine with sinus Area relation
$$\frac{c-2b \cos(A)}{b \sin(A) }=\frac{a^2-b^2}{2.Area(ABC) }$$ $$\frac{a-2c \cos(B)}{c \sin(B) }=\frac{b^2-c^2}{2.Area(ABC)}$$ $$\frac{b-2a \cos(C)}{a \sin(C) }=\frac{c^2-a^2}{2.Area(ABC)}$$
Step 6:
Add them
$$\frac{c-2b \cos(A)}{b \sin(A) }+\frac{a-2c \cos(B)}{c \sin(B) }+\frac{b-2a \cos(C)}{a \sin(C) }=\frac{a^2-b^2}{2.Area(ABC) }+\frac{b^2-c^2}{2.Area(ABC)}+\frac{c^2-a^2}{2.Area(ABC)}$$
$$\frac{c-2b \cos(A)}{b \sin(A) }+\frac{a-2c \cos(B)}{c \sin(B) }+\frac{b-2a \cos(C)}{a \sin(C) }=\frac{a^2-b^2+b^2-c^2+c^2-a^2}{2.Area(ABC) }$$
$$\frac{c-2b \cos(A)}{b \sin(A) }+\frac{a-2c \cos(B)}{c \sin(B) }+\frac{b-2a \cos(C)}{a \sin(C) }=0$$
