1）這題我沒有用迭代法
a+1/b=1,通分得ab+1=b,ab=b-1……①
b+1/c=1,通分得bc+1=c,bc=c-1……②
①×②,得 ab²c=(b-1)(c-1)
ab²c=bc-b-c+1
ab²c=(c-1)-b-(c-1)
ab²c=-b
∵b≠0
∴abc=-1……③
③代入①,得ab=b+abc
因為 b≠0
∴a=ac+1
∵a≠0
兩邊同除以a
得 c+(1/a)=1
2）x²-y²-z²=0
y²=x²-z²,z²=x²-y²
x³-y³-z³
=(x³-y³)-z(x²-y²)
=(x-y)(x²+xy+y²)-z(x+y)(x-y)
=(x-y)(x²+xy+y²-zx-zy)
=(x-y)[x²+xy+(x²-z²)-zx-zy]
=(x-y)[(x²-xz)+(x+z)(x-z)+(xy-zy)]
=(x-y)[x(x-z)+(x+z)(x-z)+y(x-z)]
=(x-y)(x-z)[x+x+z+y]
=(x-y)(x-z)(2x+y+z)
所以A=2x+y+z