若x+y+z=0,则x³+y³+z³=?

x+y+z=0
则z=-(x+y)
z³=-(x+y)³=-(x³+3x²y+3xy²+y³)=-(x³+y³)-3xy(x+y)=-(x³+y³)+3xyz
x³+y³+z³=3xyz

我的式子中 ^2,^3分别表示2次方和3次方
x^3+y^3=(x+y)(x^2-xy+y^2)
=(x+y)[(x+y)^2-3xy]
=-z[(-z)^2-3xy]
=3xyz-z^3
所以
x³+y³+z³=3xyz

x+y+z=0
=> x+y=-z
=> (x+y)^3=-z^3
=> x^3+y^3+3xy^2+3yx^2=-z^3
=> x^3+y^3+z^3+3xy^2+3yx^2=0
=> x^3+y^3+z^3+3xy^2+3yx^2+3xyz-3xyz=0
=> x^3+y^3+z^3+3xy(x+y+z)-3xyz=0 由于x+y+z=0
=> x^3+y^3+z^3=3xyz

x^3+y^3+z^3-3xyz
=[(x+y)^3-3xy(x+y)]+z^3-3xyz
=[(x+y)^3+z^3]-3xy(x+y+z)
=[(x+y+z)^3-3(x+y)z(x+y+z)]-3xy(x+y+z)
=(x+y+z)^3-3(xy+yz+zx)(x+y+z)
=(x+y+z)[(x+y+z)^2-3(xy+yz+zx)]
=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
x+y+z=0
所以x^3+y^3+z^3-3xyz=0
所以x^3+y^3+z^3=3xyz

x=-y-z
y=-x-z
z=-x-y
把右方的三方相加就是了

将 x^3+y^3+z^