Question 552011
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Factor out *[tex \Large 2x] leaving you with:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\left(125x^3\ -\ 27\right)]


And since *[tex \Large 125x^3\ =\ \left(5x\right)^3] and *[tex \Large 27\ =\ 3^3], just follow the pattern for the difference of two cubes.


BTW, the cubes factorizations are much easier to remember as one statement rather than two as you gave them:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^3\ \pm\ y^3\ =\ (x\ \pm\ y)\left(x^2\ \mp\ 2xy\ +\ y^2)]


And the key to remembering the way the signs go is to remember San Diego Padres -- SDP -- Same, Different, Positive.


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