f(x)=（x^3-2）/[2(x-1)^2]求导

f(x)=（x^3-2）/[2(x-1)^2]求导
f'(x)=[6x²(x-1)²-4(x³-2)(x-1)]/[2(x-1)²]²>0即 6x²(x-1)²-4(x³-2)(x-1)>0 (x-1)[6x²(x-1)-4(x³-2)]>0 (x-1)[6x³-6x²-4x³+8]>0 (x-1)(x³-3x²+4)>0 (x-1)[x³+x²-4x²+4]>0 (x-1)[x²(x+1)-4(x-1)(x+1)]>0 (x-1)(x+1)(x²-4+4)>0 (x-1)>0
我想问从(x-1)(x³-3x²+4)>0想到 (x-1)[x³+x²-4x²+4]>0到底是怎么想出来的?
用了什么思想?或者说x^3-3x^2+4x怎么化成(x+1)(x-2)²