Question

已知：x+y+z=0,求证：x^3+y^3+z^3=3xyz

Answer 1

因式分解
x^3+y^3+z^3-3xyz=（x+y+z）(x^2+y^2+z^2-xy-yz-zx)=0，所以原式成立

Answer 2

a^2+b^2+c^2+2ab+2ac+2bc

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

Answer 3

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
3xyz=3xy(-x-y)=-3x^2y-3xy^2
x^3+y^3+z^3=3xyz