Question 606795
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 100\ -\ c^6]


Factor the difference of two squares:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (10\ +\ c^3)(10\ -\ c^3)]


Bearing in mind that the cube root of 10 cubed is 10, factor the sum and difference of two cubes:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\sqrt[3]{10}\ +\ c\right) \left(\sqrt[3]{100}\ -\ c\sqrt[3]{10}\ +\ c^2\right)\left(\sqrt[3]{10}\ -\ c\right)\left(\sqrt[3]{100}\ +\ c\sqrt[3]{10}\ +\ c^2\right)]


You can also apply the quadratic formula to the two trinomials and get the following result:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\sqrt[3]{10}\ +\ c\right) \left(c\ -\ \frac{c\sqrt[3]{10}\ +\ \sqrt{c^2\sqrt[3]{100}\ -\ 4\sqrt[3]{100}}}{2}\right)\left(c\ -\ \frac{c\sqrt[3]{10}\ -\ \sqrt{c^2\sqrt[3]{100}\ -\ 4\sqrt[3]{100}}}{2}\right)\cdots\\ \ \ \ \ \ \ \ \ \ \ \cdots \left(\sqrt[3]{10}\ -\ c\right) \left(c\ -\ \frac{-c\sqrt[3]{10}\ +\ \sqrt{c^2\sqrt[3]{100}\ -\ 4\sqrt[3]{100}}}{2}\right)\left(c\ -\ \frac{-c\sqrt[3]{10}\ -\ \sqrt{c^2\sqrt[3]{100}\ -\ 4\sqrt[3]{100}}}{2}\right)]


Which is the end because you started with a 6th degree polynomial and ended with 6 first degree binomial factors.


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