Question

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

Answer 1

x^3+y^3+3x^2y+3xy^2+y^3+z^3+3y^2z+3yz^2+z^3+x^3+3x^2z+3z^2x_3 (xy+xz+y^2+yz)(z+x)
=2x^3+2y^3+2z^3+3 (x^2y+y^2z)+3 (xy^2+yz^2)+3 (z^2x+zx^2)-3 (xyz+x^2y+xz^2x^2z+y^2x+y^2z+xyz)
=2 (x^3+y^3+z^3)+3 (x^2y+y^2z+xy^2+yz^2+z^2x+zx^2)-3(2xyz+x^2y+xz^2+x^2z+y^2z+y^2x)
=2 (x^3+y^3+z^3)_6xyz
=2 (x^3+y^3+z^3-3xyz)

Answer 2

Step-by-step explanation:

here you go hope you understood