Question 248316
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No maybe about it.  *[tex \LARGE x\ =\ x^1].  So what do I need to multiply *[tex \LARGE x^{\frac{2}{3}}] by in order to get *[tex \LARGE x^1]?  Well, *[tex \LARGE 1\ -\ \frac{2}{3}\ =\ \frac{1}{3}] so *[tex \LARGE x^{\frac{2}{3}}\,\cdot\,x^{\frac{1}{3}}\ =\ x^{\frac{2}{3}+\frac{1}{3}} =\ x^1]



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