Question 313031
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8x^3\ +\ 343]


Is the sum of two cubes, just follow the pattern:


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



Let *[tex \Large u\ =\ x^2]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^4\ -\ 20x^2\ +\ 64\ =\ u^2\ -\ 20u\ +\ 64]


Factor the quadratic in *[tex \Large u], then substitute back *[tex \Large x^2\ =\ u] and factor each of the factors.  You will end up with 4 factors which is what we expected from a 4th degree polynomial.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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