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Sir pls solve the following.
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Asked by rsudipto | 28 Dec, 2018, 07:54: PM
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begin mathsize 16px style 1 plus straight omega plus straight omega squared equals 0 straight omega cubed equals 1......... left parenthesis straight i right parenthesis straight omega to the power of 3 straight n end exponent equals 1 straight omega to the power of 4 equals straight omega cubed space straight omega space equals space straight omega....... left parenthesis ii right parenthesis straight S equals fraction numerator straight a plus bω plus cω squared over denominator straight b plus cω plus aω squared end fraction plus fraction numerator straight a plus bω plus cω squared over denominator straight c plus aω plus bω squared end fraction straight S equals straight omega over straight omega fraction numerator straight a plus bω plus cω squared over denominator straight b plus cω plus aω squared end fraction plus straight omega squared over straight omega squared fraction numerator straight a plus bω plus cω squared over denominator straight c plus aω plus bω squared end fraction straight S equals fraction numerator straight omega open parentheses straight a plus bω plus cω squared close parentheses over denominator straight b straight omega plus cω squared plus aω cubed end fraction plus fraction numerator straight omega squared open parentheses straight a plus bω plus cω squared close parentheses over denominator cω plus aω squared plus bω cubed end fraction Simplifying space using space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis space we space get straight S equals straight omega plus straight omega squared 1 plus straight omega plus straight omega squared equals 0 straight omega plus straight omega squared equals straight S equals negative 1 end style
Answered by Sneha shidid | 31 Dec, 2018, 12:08: PM

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