1、(x-1)(x-4)=x²-5x+4
(x-2)(x-3)=x²-5x+6
令t=x²-5x+4
原式=t(t+2)-120=t²+2t-120=(t+12)(t-10)
把t=x²-5x+4代入得：原式=(x²-5x+16)(x²-5x-6)=(x²-5x+16)(x-6)(x+1)
2、x8+x6+x4+x2+1=(x10-1)/(x2-1)=(x5+1)(x5-1)/(x+1)(x-1)
在已知公式中令b=1可求出(a5-1)/(a-1)和(a5+1)/(a+1)的结果