Question 351646
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Presumeably you want a complete factorization of this 6th degree polynomial expression.


Let *[tex \Large u\ =\ y^3].


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2y^6\ -\ 128\ =\ 2u^2\ -\ 128]


Take out a 2:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\left(u^2\ -\ 64\right)]


Factor the quadratic binomial:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\left(u\ -\ 8\right)\left(u\ +\ 8\right)]


Replace the substitution:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\left(y^3\ -\ 8\right)\left(y^3\ +\ 8\right)]


Use the sum and difference of cubes factorization:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\left(y\ -\ 2\right)\left(y^2\ +\ 2y\ +\ 4\right)\left(y\ +\ 2\right)\left(y^2\ -\ 2y\ +\ 4\right)]


If you are factoring over *[tex \Large \mathbb{R}], then you are done.  If you need to factor over *[tex \Large \mathbb{C}], then continue using the quadratic formula to determine the four complex factors (verification left as an exercise for the student):


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\left(y\ -\ 2\right)\left(y\ +\ 1\ -\ i\sqrt{3}\right)\left(y\ +\ 1\ +\ i\sqrt{3}\right)\left(y\ +\ 2\right)\left(y\ -\ 1\ -\ i\sqrt{3}\right)\left(y\ -\ 1\ +\ i\sqrt{3}\right)]


Ending with 6 binomial factors as the Fundamental Theorem of Algebra guarantees for a 6th degree polynomial.


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