已知x^1/3+y^1/3+z^1/3=0 求(x+y+z)³-27xyz的值?

已知x^(1/3)+y^(1/3)+z^(1/3)=0 求(x+y+z)^3－27xyz的值?
Sol
Set a=x^(1/3)，b=y^(1/3)，c=z^(1/3)
a+b+c=0
c=－a－b
c^3=－a^3－3a^2b－3ab^2－b^3
a^3+b^3+c^3
=a^3+b^3－a^3－3a^2b－3ab^2－b^3
=－3a^2b－3ab^2
A=(x+y+z)^3－27xyz
=(a^3+b^3+c^3)^3－27a^3b^3c^3
=(－3a^2b-3ab^2)^3－27a^3b^3c^3
A/(27a^3b^3)
=－(a+b)^3-c^3
=－(a+b)^3+(a+b)^3
=0
(x+y+z)^3－27xyz=0