Question 641922
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 48k^4\ -\ 243]


Note: *[tex \LARGE 48\ = 3\,\cdot\,16\ =\ 3\,\cdot\,2^4]  and *[tex \LARGE 243\ =\ 3^5], so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 48k^4\ -\ 243\ =\ 3\left((2k)^4\ -\ 3^4\right)]


Which is the difference of two squares, so:


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


Then we have the difference of two squares again, so:


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


And if you are factoring over the Reals, you are done.  If you are factoring over the Complex numbers, you can go one more step because *[tex \LARGE (2k)^2\ +\ 3^2\ =\ (2k)^2\ -\ (3i)^2] where *[tex \LARGE i^2\ =\ -1], and this is also the difference of two squares, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3(2k\ -\ 3)(2k\ +\ 3)(2k\ -\ 3i)(2k\ +\ 3i)]


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