令t=x-1即x=t+1
则f(t)=(t+1)^3-3(t+1)^2+2(t+1)=(t+1)[(t+1)^2-2(t+1)-t+1]=(t+1)[(t+1-1)^2-t]=(t+1)(t-1)t
当t=2x时,即f(2x)=2x(2x+1)(2x-1)=8x^3-2x
令t=x-1即x=t+1
则f(t)=(t+1)^3-3(t+1)^2+2(t+1)=(t+1)[(t+1)^2-2(t+1)-t+1]=(t+1)[(t+1-1)^2-t]=(t+1)(t-1)t
当t=2x时,即f(2x)=2x(2x+1)(2x-1)=8x^3-2x