<PRE><B>Question 640265</B>
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(2\ +\ h)^3\ -\ 2^3}{h}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{8\ +\ 12h\ +\ 6h^2\ +\ h^3\ -\ 8}{h}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{12h\ +\ 6h^2\ +\ h^3}{h}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 12\ +\ 6h\ +\ h^2]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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