证明:由于x+y+z=0,故有z=-x-y
左边=x^3+y^3+z^3=x^3+y^3+(-x-y)^3
=x^3+y^3-(x+y)^3
=x^3+y^3-(x^3+3x^2y+3xy^2+y^3)
=x^3+y^3-x^3-3x^2y-3xy^2-y^3=-3x^2y-3xy^2
右边=3xyz=3xy(-x-y)=-3x^2y-3xy^2
所以有:x^3+y^3+z^3=3xyz
证明:由于x+y+z=0,故有z=-x-y
左边=x^3+y^3+z^3=x^3+y^3+(-x-y)^3
=x^3+y^3-(x+y)^3
=x^3+y^3-(x^3+3x^2y+3xy^2+y^3)
=x^3+y^3-x^3-3x^2y-3xy^2-y^3=-3x^2y-3xy^2
右边=3xyz=3xy(-x-y)=-3x^2y-3xy^2
所以有:x^3+y^3+z^3=3xyz