f'（x）=12ax^3-12ax-4x+4
=12ax(x^2-1)-4(x-1)
=12ax(x+1)(x-1)-4(x-1)
=4（x-1)(3ax^2+3ax-1）
当a=1/6时,f'（x）= 2（x-1)*2(1/2x^2+1/2x-1)
= 2(x-1)(x^2+x-2)
= 2(x+2)(x-1)^2
帮忙解一下由题干怎么到划线的第一步,第一步又是怎么到划线第二步的
f'（x）=12ax^3-12ax-4x+4
=12ax(x^2-1)-4(x-1)
=12ax(x+1)(x-1)-4(x-1)
=4（x-1)(3ax^2+3ax-1）
当a=1/6时,f'（x）= 2（x-1)*2(1/2x^2+1/2x-1)
= 2(x-1)(x^2+x-2)
= 2(x+2)(x-1)^2