Question

已知[a+(a-1)分之1][(1+a的3次方)分之1 -1]除以(1-a)分之a的3次方,其中a=99,求原式的值

Answer 1

[a+(a-1)分之1][(1+a的3次方)分之1 -1]除以(1-a)分之a的3次方
=(a-1)分之(a²-a+1) × (1+a³)分之(-a³) × a³分之(1-a)
= (a²-a+1) × [(a+1)(a²-a+1)]分之1
= (a+1)分之1
所以当a=99时，原式=(99+1)分之1=100分之1

Answer 2

[a+(a-1)分之1][(1+a的3次方)分之1 -1]除以(1-a)分之a的3次方
=﹛[﹙a²-a+1﹚／﹙a-1﹚]×[﹙-a³﹚／﹙a³+1﹚]﹜／[﹙a³﹚／﹙a-1﹚]
=1/﹙a+1﹚
=1/100