\(P=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}-\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\\ =\dfrac{ab\left(a+b\right)}{abc}+\dfrac{bc\left(b+c\right)}{abc}+\dfrac{ca\left(c+a\right)}{abc}-\left(\dfrac{abc}{a}+\dfrac{abc}{b}\right)\left(\dfrac{abc}{b}+\dfrac{abc}{c}\right)\left(\dfrac{abc}{c}+\dfrac{abc}{a}\right)\\ =a^2b+ab^2+b^2c+bc^2+c^2a+ca^2-\left(bc+ca\right)\left(ca+ab\right)\left(ab+bc\right)\\ =ab\left(a+b\right)+c\left(a^2+b^2\right)+c^2\left(a+b\right)-abc\left(a+b\right)\left(b+c\right)\left(c+a\right)\\ =ab\left(a+b\right)+c\left(a+b\right)^2-2abc+c^2\left(a+b\right)-\left(a+b\right)\left(b+c\right)\left(c+a\right)\\ =\left(a+b\right)\left[ab+c\left(a+b\right)+c^2\right]-\left(a+b\right)\left(b+c\right)\left(c+a\right)-2\\ =\left(a+b\right)\left(b+c\right)\left(c+a\right)-\left(a+b\right)\left(b+c\right)\left(c+a\right)-2\\ =-2\)

#$\mathtt{Toru}$

\(a\text{) }\dfrac{x^2-3x+2}{x+3}=\dfrac{x^2+3x-6x-18+20}{x+3}\\ =\dfrac{x\left(x+3\right)-6\left(x+3\right)+20}{x+3}=x-6+\dfrac{20}{x+3};\left(x\ne-3\right)\\ b\text{) }\dfrac{2x^2-x+4}{x-2}=\dfrac{2x^2-4x+3x-6+10}{x-2}\\ =\dfrac{2x\left(x-2\right)+3\left(x-2\right)+10}{x-2}\\ =2x+3+\dfrac{10}{x-2};\left(x\ne2\right)\)

\(c\text{) }\dfrac{x^2}{2x-4}=\dfrac{x^2-2x+2x-4+4}{2\left(x-2\right)}\\ =\dfrac{x\left(x-2\right)+2\left(x-2\right)+4}{2\left(x-2\right)}=x+2+\dfrac{2}{x-2};\left(x\ne2\right)\\ d\text{) }\dfrac{3x^2-2x+1}{2-x}=-\dfrac{3x^2-6x+4x-8+9}{x-2}\\ =-\dfrac{3x\left(x-2\right)+4\left(x-2\right)+9}{x-2}=-3x-4-\dfrac{9}{x-2};\left(x\ne2\right)\)

#$\mathtt{Toru}$