Question

ANY ONE PLZ HELP 25 POINTS PROVE THAT x³+y³+z³-3xyz=(x+y+z)(x+ωy+ω²z)(x+ω²y+ωz)

Answer 1

Proved

Explanation:

Taking R.H.S

=>
`(x+y+z)[x(x+w^2y+wz)+wy(x+w^2y+wz)+w^2z(x+w^2y+wz)]`

=>
`(x+y+z)(x^2+w^2xy+wxz+wxy+w^3y^2+w^2yz+w^2xz+w^4yz+w^3z^2)`

Remember ∴ ω³ = 1

So we'll replace all ω³'s with 1

=>
`(x+y+z)(x^2+y^2+z^2+w^2xy+wxy+w^2yz+w^4yz+w^2xz+wxz)`

=>
`(x+y+z)[x^2+y^2+z^2+xy(w^2+x)+w^2yz+w^3*wyz+xz(w^2+w)]`

Remember ∴ ω²+ω = -1

=>
`(x+y+z)[x^2+y^2+z^2+xy(-1)+yz(w^2+w)+xz(-1)]`

=>
`(x+y+z)(x^2+y^2+z^2-xy-yz-xz)`

According to formula:

x³+y³+z³-3xyz =
`(x+y+z)(x^2+y^2+z^2-xy-yz-xz)`

So, it becomes

=>
`x^3+y^3+z^3-3xyz`