Question

If 1, w,ware cube roots of unity, then prove that:
1. (1+w^2)^4 = w​

Answer 1

Given:

$\star \bold{ 1 . \: \: \omega \: \: \: \: and \: \: { \omega}^{2} \: are \: cube }\: \: \\ \bold{root \ \: of \: unity \: }.$

To prove:

$\star \bold{( {1 + { \omega}^{2}) }^{4} = \omega}$

Solution:

• As we know that ,

$\star \bold{ \boxed{1 + \omega + { \omega}^{2} = 0}} \\ \\ and \\ \\ \star \bold{ \boxed{ { \omega}^{3} = 1}}$

$\implies \: (1 + { \omega}^{2} ) ^{4} \\ \\ \implies \: ( - \omega) ^{4} \: \: \: \: \: \: \: \boxed {1 + { \omega}^{2} = - \omega} \\ \\ \implies \: { \omega}^{4} \\ \\ \implies \: { \omega}^{3} \omega \: \\ \\ \implies \: 1 (\omega) \: \: \: \: \: \: \: \: \: \boxed{ \red{\omega ^{3} = 1}} \\ \\ \implies \: \omega$

$\huge \green{ \mathfrak{proved}}$

Answer 2

Step-by-step explanation:

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