Question 1123448
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Let *[tex \Large u\ =\ z^3], then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u^2\ +\ (2\ +\ 2i)u\ +\ 2i]


Using the Quadratic formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ -(1\ +\ i)]


with a multiplicity of 2.  Verification of this step left as an exercise for the student.


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z^3\ =\ -1\ -\ i]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z\ =\ \sqrt[3]{-1\ -\ i}]


But this must be multiplied by the three cube roots of unity to obtain the three roots of the cubic which, considering the multiplicity of the quadratic in *[tex \Large u] gives the 6 roots of the original 6th-degree equation.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z_{1,2}\ =\ \sqrt[3]{-1\ -\ i}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z_{3,4}\ =\ -\sqrt[3]{-1}\,\cdot\,\sqrt[3]{-1\ -\ i}\ =\ -\sqrt[3]{1\ +\ i}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z_{5,6}\ =\ (-1)^{\small{2/3}}\,\cdot\,\sqrt[3]{-1\ -\ i}]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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