Question 986554
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (a\ -\ b)^n\ =\ \sum_{k=0}^n\ (-1)^k{{n}\choose{k}}\,a^{n-k}b^k]


*[tex \Large {{n}\choose{k}}\ =\ \frac{n!}{k!(n-k)!}], *[tex \Large a\ =\ 4x], *[tex \Large b\ =\ 3y], and, since *[tex \Large k] starts at zero, for the fifth term, you want *[tex \Large k\ =\ 4]


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \