Question 399468
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Use the difference of two cubes factorization:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^3\ -\ b^3\ =\ (a\ -\ b)(a^2\ +\ ab\ +\ b^2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z^3\ -\ 1\ =\ (z\ -\ 1)(z^2\ +\ z\ +\ 1)]


One root is 1, but in complex form that would be *[tex \Large z\ =\ 1\ +\ 0i].  Therefore, plot the point (1,0) on your *[tex \Large x,yi] axes.


For the other two roots, solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z^2\ +\ z\ +\ 1\ =\ 0]


using the quadratic formula to get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z\ =\ \frac{-1\ \pm\ i sqrt{3}}{2}]


Hence, plot *[tex \Large \left(-\frac{1}{2},\frac{i\sqrt{3}}{2}\right)] and *[tex \Large \left(-\frac{1}{2},-\frac{i\sqrt{3}}{2}\right)] on your *[tex \Large x,yi] axes.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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