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

Question

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

1 Answer

Impact of this question

44185 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-18T22:36:02

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)#

Science

Math

Humanities

... and beyond

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

1 Answer

Related questions

Impact of this question