Ta có:
\(+)\frac{2bc-2016}{3c-2bc+2016}=-1+\frac{3c}{3c-2bc+2016}\left(1\right)\)
\(+)\frac{-2b}{3-2b+ab}=\frac{-2bc}{3c-2bc+abc}=\frac{-2bc}{3c-2bc+2016}\left(2\right)\)
\(+)\frac{4032-3ac}{3ac-4032+2016a}=-1+\frac{2016a}{3ac-2abc+2016a}=-1+\frac{2016}{3c-2bc+2016}\left(3\right)\)
\(P=\left(1\right)+\left(2\right)+\left(3\right)=-1\)
Vậy .........