a=1/(1-√2)=-1/(√2-1)=-(√2+1)/[(√2+1)(√2-1)]=-√2-1
b=1/(1+√2)=(√2-1)/[(√2+1)(√2-1)]=√2-1
a+b=-√2-1+√2-1=-2
ab=-(√2+1)(√2-1)=-1
a³b+ab³
=ab(a²+b²)
=ab[(a+b)²-2ab]
=(-1)[(-2)²-2(-1)]
=-6
分子有理化
a=1/(1-√2) =(1+√2)/(1-√2)(1+√2) =-(1+√2)
b=1/(1+√2) =(1-√2)/(1+√2)(1-√2) =-(1-√2)
a³b+ab³=ab(a²+b²)
a²+b²=(1+√2)²+(1-√2)²=2×(1+2)=6
ab=(1+√2)×(1-√2)=1-2=-1
所以a³b+ab³=6×(-1)=-6