How do you factor completely x^3+y^3+z^3-3xyz x3+y3+z3−3xyz?

1 Answer
Konstantinos Michailidis
Nov 18, 2015

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)

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