Question 276153
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I'm not sure who "they" are, but "they" certainly have a different system of mathematics than I ever saw.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{7x}{\sqrt[3]{2x^2y}}\ =\ \frac{7x}{\sqrt[3]{2x^2y}}\left(\frac{\sqrt[3]{2x^2y}}{\sqrt[3]{2x^2y}}\right)^2\ =\ \frac{7x\sqrt[3]{4x^4y^2}}{2x^2y}\ =\ \frac{7x^2\,\sqrt[3]{4xy^2}}{2x^2y}\ =\ \frac{7\sqrt[3]{4xy^2}}{2y}]


That's my story, an' I'm stickin' to it.


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