令x+2=t,则x+1=t-1,x+3=t+1(x+1)(x+2)(x+3)-6×7×8=(t-1)t(t+1)-(7-1)×7×(7+1)=t(t^2-1)-7×(7^2-1)=t^3-t-7^3+7=(t^3-7^3)-(t-7)=(t-7)(t^2+7t+49)-(t-7)=(t-7)(t^2+7t+48)=(x+2-7)[(x+2)^2+7(x+2)+48]=(x-5)(x^...