Question

已知a^(2/3)+b^(2/3)=4,x＝a＋3a^(1/3)·b^(2/3)

y＝b＋3a^(2/3)·b^(1/3),求证：（x＋y）^(2/3)＋（x－y）^(2/3)为定值

Answer 1

设a=8sinθ^3,b=8cosθ^3
代入，得x＝a＋3a^(1/3)·b^(2/3)=8sinθ^3+24sinθcosθ^2
y=8cosθ^3+24sinθ^2cosθ
则x+y
=8sinθ^3+24sinθcosθ^2+8cosθ^3+24sinθ^2cosθ
=8sinθ(sinθ^2+3cosθ^2)+8cosθ(cosθ^2+3sinθ^2)
=8sinθ(1+2cosθ^2)+8cosθ(1+2sinθ^2)
=8(cosθ+sinθ)+16sinθcosθ(cosθ+sinθ)=8(1+2sinθcosθ)(sinθ+cosθ)
=8(sinθ^2+2sinθcosθ+cosθ^2）(sinθ+cosθ)=8(sinθ+cosθ)^3
同理，x-y=8(sinθ-cosθ)^3
则（x＋y）^(2/3)＋（x－y）^(2/3)=4(sinθ+cosθ)^2+4（sinθ-cosθ)^2
=4*2（sinθ^2+cosθ^2)=8为定值