Question

prove that (1+omega-omega²)(1-omega+omega²)=4​

Answer 1

Step-by-step explanation:

(1+ω-ω²)(1-ω+ω²)

=(-ω-ω²+ω-ω²)(-ω-ω²-ω+ω²) [1+ω+ω²=0]

=(-2ω²)(-2ω)

=4ω³

=4

Answer 2

Answer:

We know

${w}^{2} + w + 1 = 0$

So

$1 + w = - {w}^{2} \: \: \: and \: \: \: 1 + {w}^{2} = - w$

Put the value

$( - {w}^{2} - {w}^{2} )( - w - w) \\ = ( - 2 {w}^{2} )( - 2w) \\ = 4 {w}^{3} \\ = 4$