Question

if 1, omega , omega² are cube roots of unity then find the value of (3+omega²+omega⁴)raise to 6​

Answer 1

Answer:

The correct answer is 64.

Step-by-step explanation:

The steps are given in the attachment.

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Answer 2

$\huge\mathfrak{solution}$

Given:

$1 \: \: \: \: \: \omega \: \: and \: \: { \omega}^{2} \: \bold{are \: the \: cube \: roots} \\ \bold{ \: of \: unity}$

To find :

$(3 + { \omega}^{2} + { \omega}^{4} ) ^{6}$

As we know that :

$\boxed{1 + \omega \: + { \omega}^{2} = 0}$

$\star {(3 + { \omega}^{2} + { \omega}^{3} \omega )}^{6} \\ \\ \star \: (3 + { \omega}^{2} + \omega) ^{6} \: \: \: \: \: \: \boxed{ { \omega}^{3} = 1} \\ \\ \star \: (3 + ( - 1))^{6} \: \: \: \: \boxed{ { \omega}^{2} + \omega = - 1} \\ \\ \star \: ({2})^{6} \\ \\ \star \: 64$

Hence the value is 64.