<s>[INST] 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?
 [/INST] Using  Proof of the sine rule 
and  Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,
$$(b^2-c^2)\cot A$$
$$=4R^2(\sin^2B-\sin^2C)\cot A$$
$$=4R^2\sin(B+C)\sin(B-C)\cdot\dfrac{-\cos(B+C)}{\sin(B+C)}$$
$$=-4R^2\sin(B-C)\cos(B+C)=2R^2(\sin2C-\sin2B)$$
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