In $\triangle ABC$, $(b^2-c^2) \cot A+(c^2-a^2) \cot B +(a^2-b^2) \cot C = $? 
In $\triangle ABC$, 
  $$(b^2-c^2) \cot A+(c^2-a^2) \cot B +(a^2-b^2) \cot C = \text{?}$$

I tried solving this by cosine rule but it is becoming too long. Any short step solution for this?
 A: Solving this problem by Law of Cosines isn't short but not that unmanageable.
The key is the observation of pattern within the expression:
$$(b^2-c^2)\cot A = \left.(b^2-c^2)\frac{b^2+c^2-a^2}{2bc}\right/\frac{a}{2R} = \frac{R}{abc} \left[ b^4 - c^4 - a^2b^2 + b^2c^2\right]$$ 
This has the form $\phi(a,b,c) - \phi(b,c,a)$ where 
$$\phi(a,b,c) = \frac{R}{abc}(b^4 - a^2b^2)$$ 
Other two terms have a similar form. They can be obtained from above by replacing $(a,b,c)$ with $(b,c,a)$ and $(c,a,b)$ respectively.
When you sum over all 3 terms, you are performing a cyclic sum over $a,b,c$.
Terms of the form $\phi(a,b,c) - \phi(b,c,a)$ will simply cancel each other.
The end result is $0$.
$$\sum_{cyc} (b^2-c^2)\cot A 
= \sum_{cyc} \left( \phi(a,b,c) - \phi(b,c,a)\right)
= \sum_{cyc}\phi(a,b,c) - \sum_{cyc}\phi(a,b,c) = 0$$
In cases when you need to present a complete derivation but you
don't want to become too verbose, you can do something like this:
Let $R$ be the circumradius, we have
$$\begin{align}
\sum_{cyc} (b^2-c^2)\cot A 
&= \sum_{cyc} \left.(b^2-c^2)\frac{b^2+c^2-a^2}{2bc}\right/\frac{a}{2R}\\
&= \frac{R}{abc} \sum_{cyc} \left[ b^4 - c^4 - a^2b^2 + b^2c^2\right]\\
&= \frac{R}{abc} \left[ \sum_{cyc} (b^4 - a^2b^2) - \sum_{cyc}(c^4 - b^2c^2)\right]\\
&=  \frac{R}{abc}\left[ \sum_{cyc} (b^4 - a^2b^2) - \sum_{cyc}(b^4 - a^2b^2)\right]\\
&= 0
\end{align}
$$ 
A: Using  Proof of the sine rule 
and  Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,
$$(b^2-c^2)\cot A$$
$$=4R^2(\sin^2B-\sin^2C)\cot A$$
$$=4R^2\sin(B+C)\sin(B-C)\cdot\dfrac{-\cos(B+C)}{\sin(B+C)}$$
$$=-4R^2\sin(B-C)\cos(B+C)=2R^2(\sin2C-\sin2B)$$
