Question 1136614
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The number under the cubic root is  {{{(27/2)*(sqrt(3) + j)}}}.


Present it in the trigonometric form.


The modulus is  {{{(27/2)*sqrt((sqrt(3))^2 + 1^2))}}} = {{{(27/2)*sqrt(4)}}} = 27.


The argument is  {{{arctan(1/sqrt(3))}}} = 30°.


Now apply the de Moivre's formula.


The third degree roots of your number are 


    1)  3*cis(10°) = {{{3*(cos(10^o) + j*sin(10^o))}}};


    2)  3*cis(130°) = {{{3*(cos(130^o) + j*sin(130^o))}}};


    3)  3*cis(250°) = {{{3*(cos(250^o) + j*sin(250^o))}}}.
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Solved.