\(a,x\ne\pm2\\ b,\\ =\dfrac{x^3-x\left(x+2\right)-2\left(x-2\right)}{x^2-4}\\ =\dfrac{x^3-x^2-2x-2x+4}{x^2-4}=\dfrac{x^3-4x-x^2+4}{x^2-4}\\ =\dfrac{x\left(x^2-4\right)-\left(x^2-4\right)}{x^2-4}=\dfrac{\left(x^2-4\right)\left(x-1\right)}{x^2-4}\\ =x-1\)