Question 576349
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*[tex \LARGE (x\ +\ 1)^3\ =\ (x\ +\ 1)^2(x\ +\ 1)\ \cr.\cr =\ (x^2\ +\ 2x\ +\ 1)(x\ +\ 1)\ \cr.\cr =\ x^3\ +\ 2x^2\ +\ x\ +\ x^2\ +\ 2x\ +\ 1\ \cr.\cr =\ x^3\ +\ 3x^2\ +\ 3x\ +\ 1]


Is the *[tex \LARGE (x\ +\ 1)^3] part.  Then just multiply the result above by each of the other two factors.  Note *[tex \LARGE x\ -\ 0] is just *[tex \LARGE x]


In general, the binomial expansion is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ y)^n\ =\ \sum_{k\,=\,0}^n\ \left(n\cr k\right)\,x^{n\,-\,k}y^k]


Where *[tex \LARGE \left(n\cr k\right\)] is the number of combinations of *[tex \LARGE n] things taken *[tex \LARGE k] at a time and is calculated by *[tex \LARGE \frac{n!}{k!(n\,-\,k)!}] 


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