Math, asked by junali007, 9 months ago

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

Answers

Answered by nipun2624
3

Answer:

The correct answer is 64.

Step-by-step explanation:

The steps are given in the attachment.

HOPE THIS WILL HELP YOU. IF SO MARK IT AS BRAINLIEST.

Attachments:
Answered by kaushik05
25

  \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.

Similar questions