已知：x+y+z=0.求证：x三次+y上次+z三次=3xyz.

x³+y³+z³-3xyz
=[(x+y)³+z³]-3xy(x+y+z)
=(x+y+z)(x²+y²+2xy-xz-yz+z²)-3xy(x+y+z)
=(x+y+z)(x²+y²+z²+2xy-3xy-xz-yz)
=(x+y+z)(x²+y²+z²-xy-yz-xz)=0
所以x³+y³+z³=3xyz