Question

If w denotes the complex cube-root of unity, prove the following
i) (w² + w - 1)³ = -8
ii) (a + b) + (aw + bw²) + (aw² + bw) = 0

Answer 1

➡️ Solution:

we know that
$1 + w + {w}^{2} = 0 \\ \\ {w}^{3} = 1$
For the Solution of

$1) \: {( {w}^{2} + w - 1) }^{3} = - 8 \\ \\ we \: know \: that \\ \\ w + {w}^{2} = - 1 \\ \\ = {( - 1 - 1) }^{3} \\ \\ = ( { - 2)}^{3} \\ \\ = - 8 \\ \\ = RHS$
By the same way

$2) \: (a + b) + (aw + b {w}^{2} ) + (a {w}^{2} + bw) = 0 \\ \\ = a + aw + a {w}^{2} + b + bw + b {w}^{2} \\ \\ = a(1 + w + {w}^{2} ) + b(1 + w + {w}^{2} ) \\ \\ = a(0) + b(0) \\ \\ = 0 \\ \\ = RHS$
Hence proved.

Hope it helps you.