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?

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
$$\begin{align}
\sum_{cyc} (b^2-c^2)\cot A 
&= \sum_{cyc} \left.(b^2-c^2)\frac{b^2+c^2-a^2}{2bc}\right/\frac{a}{2R}\\
&= \frac{R}{abc} \sum_{cyc} \left[ b^4 - c^4 - a^2b^2 + b^2c^2\right]\\
&= \frac{R}{abc} \left[ \sum_{cyc} (b^4 - a^2b^2) - \sum_{cyc}(c^4 - b^2c^2)\right]\\
&=  \frac{R}{abc}\left[ \sum_{cyc} (b^4 - a^2b^2) - \sum_{cyc}(b^4 - a^2b^2)\right]\\
&= 0
\end{align}
$$ 
