分解因式:(1-7t-7t2-3t3)(1-2t-2t2-t3)-(t+1)6=________.
t(2t2+5t+5)(t-1)(t2+2t+3)
分析:可設(t+1)3=x,y=t2+t+2,將(1-7t-7t2-3t3)(1-2t-2t2-t3)-(t+1)6變形為(2y-3x)(y-x)-x2的形式分解因式.
解答:設(t+1)3=x,y=t2+t+2,則
原式=[2(t2+t+2)-3(1+3t+3t2+t3)][(t2+t+2)-(1+3t+3t2+t3)]-[(t+1)3]2
=(2y-3x)(y-x)-x2
=2x2-5xy+2y2
=(2x-y)(x-2y)
=[2(t3+3t2+3t+1)-(t2+t+2)][(t3+3t2+3t+1)-2(t2+t+2)]
=(2t3+5t2+5t)(t3+t2+t-3)
=t(2t2+5t+5)(t-1)(t2+2t+3).
故答案為:t(2t2+5t+5)(t-1)(t2+2t+3).
點評:本題考查了用換元法分解因式,它能夠把一些整式化繁為簡,化難為易,對此應注意總結能用換元法分解因式的特點,尋找解題技巧.