Question 326146
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 36x^2\ =\ (6x)^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ =\ 1^2]


So you have the difference of two squares:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^2\ -\ b^2\ =\ (a\ +\ b)(a\ -\ b)]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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