\(\dfrac{ax^3+bx^2+c}{x-2}=\dfrac{ax^3-2ax^2+\left(b+2a\right)x^2-2\left(b+2a\right)x+2\left(b+2a\right)x-4\left(b+2a\right)+4b+8a+c}{x-2}\)
=>8a+4b+c=0
\(\dfrac{ax^3+bx^2+c}{x^2-1}\)
\(=\dfrac{ax^3-ax+bx^2-b+ax+b+c}{x^2-1}\)
\(=ax+b+\dfrac{ax+b+c}{x^2-1}\)
=>ax+b+c=2x+5
=>a=2; b+c=5;
8a+4b+c=0
=>4b+c+16=0
=>4b+c=-16
=>b=-7; c=12