Answer:
Proved
Step-by-step explanation:
Taking R.H.S
=> [tex](x+y+z)[x(x+w^2y+wz)+wy(x+w^2y+wz)+w^2z(x+w^2y+wz)][/tex]
=> [tex](x+y+z)(x^2+w^2xy+wxz+wxy+w^3y^2+w^2yz+w^2xz+w^4yz+w^3z^2)[/tex]
Remember ∴ ω³ = 1
So we'll replace all ω³'s with 1
=> [tex](x+y+z)(x^2+y^2+z^2+w^2xy+wxy+w^2yz+w^4yz+w^2xz+wxz)[/tex]
=> [tex](x+y+z)[x^2+y^2+z^2+xy(w^2+x)+w^2yz+w^3*wyz+xz(w^2+w)][/tex]
Remember ∴ ω²+ω = -1
=> [tex](x+y+z)[x^2+y^2+z^2+xy(-1)+yz(w^2+w)+xz(-1)][/tex]
=> [tex](x+y+z)(x^2+y^2+z^2-xy-yz-xz)[/tex]
According to formula:
x³+y³+z³-3xyz = [tex](x+y+z)(x^2+y^2+z^2-xy-yz-xz)[/tex]
So, it becomes
=> [tex]x^3+y^3+z^3-3xyz[/tex]
