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Asked by g_archanasharma | 18 Dec, 2018, 19:31: PM
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begin mathsize 12px style open parentheses x minus 3 close parentheses cubed plus 1 space equals space 0 L e t space y space equals space x minus 3 space semicolon space t h e n space w e space h a v e space space space y cubed plus 1 space equals space 0 y cubed space equals space minus 1 space equals space cos left parenthesis 2 k plus 1 right parenthesis straight pi space plus space straight i space sin open parentheses 2 straight k plus 1 close parentheses straight pi space space comma space straight k space equals space 0 comma 1 comma 2 straight y space equals space open parentheses cos left parenthesis 2 straight k plus 1 right parenthesis straight pi space plus space straight i space sin open parentheses 2 straight k plus 1 close parentheses straight pi close parentheses to the power of bevelled 1 third end exponent space equals space open parentheses cos space fraction numerator left parenthesis 2 straight k plus 1 right parenthesis straight pi over denominator 3 end fraction space plus space straight i space sin space fraction numerator left parenthesis 2 straight k plus 1 right parenthesis straight pi over denominator 3 end fraction close parentheses space comma space straight k space equals 0 comma space 1 comma space 2 when space straight k space equals space 0 comma space straight y space equals space open parentheses cos space straight pi over 3 space plus space straight i space sin space straight pi over 3 close parentheses space equals space 1 half plus straight i fraction numerator square root of 3 over denominator 2 end fraction when space straight k space equals 1 comma space straight y space equals space open parentheses cos space straight pi space plus space straight i space sin space straight pi close parentheses space equals space minus 1 when space straight k equals 2 comma space straight y space equals space open parentheses cos space fraction numerator 5 straight pi over denominator 3 end fraction space plus space straight i space sin space fraction numerator 5 straight pi over denominator 3 end fraction close parentheses space equals space 1 half minus straight i fraction numerator square root of 3 over denominator 2 end fraction end style
 
since y = (x-3), sum of complex roots  = begin mathsize 12px style 3 plus 1 half plus i fraction numerator square root of 3 over denominator 2 end fraction plus 3 plus 1 half minus i fraction numerator square root of 3 over denominator 2 end fraction space equals space 7 end style
Answered by Thiyagarajan K | 19 Dec, 2018, 13:49: PM

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