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^2+4x+4+7x+14+48)
=(x-5)(x^2+11x+66)
（x+1）（x+2）（x+3）-6*7*8用换元法分解因式