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

${x}^{3} + {y}^{3} + {z}^{3} - 3 x y z = {x}^{3} + {y}^{3} + 3 {x}^{2} y + 3 x {y}^{2} + {z}^{3} - 3 x y z - 3 {x}^{2} y - 3 x {y}^{2} = {\left(x + y\right)}^{3} + {z}^{3} - 3 x y \left(x + y + z\right) = \left(x + y + z\right) \left({\left(x + y\right)}^{2} + {z}^{2} - \left(x + y\right) z\right) - 3 x y \left(x + y + z\right) = \left(x + y + z\right) \left({x}^{2} + 2 x y + {y}^{2} + {z}^{2} - x y - x z - 3 x y\right) = \left(x + y + z\right) \left({x}^{2} + {y}^{2} + {z}^{2} - x y - y z - z x\right)$