Question

How do you factor completely `x^3+y^3+z^3-3xyz`?

Answer 1

It is

`x^3+y^3+z^3-3xyz=x^3+y^3+3x^2y+3xy^2+z^3-3xyz-3x^2y-3xy^2=(x+y)^3+z^3-3xy(x+y+z)=(x+y+z)((x+y)^2+z^2-(x+y)z)-3xy(x+y+z)=(x+y+z)(x^2+2xy+y^2+z^2-xy-xz-3xy)=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)`