Question 348034
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The quick and easy way to do this one is to recognize that you have a pair of conjugates if you consider *[tex \Large x\ +\ y] as a single term and 1 as the other term.


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ \left((x\ +\ y)\ +\ 1\right)\left((x\ +\ y)\ -\ 1\right)\ =\ \left(x\ +\ y\right)^2\ -\ 1]


Then expand the binomial to get:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ \left(x\ +\ y\right)^2\ -\ 1\ =\ x^2\ +\ 2xy\ +\ y^2\ -\ 1]



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