since it can also be written as x^(1/3) and therefore 1/(x^3) and this would not make sense for x=0 because of the division with 0. So why is 0 in the domain?
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4because in most of all cases $x^{1/3}\ne \frac1{x^3}$. And because obviously $0^3=0$ (similary, $0$ is also in the domain of the square root function) – Hagen von Eitzen Jan 22 '20 at 17:34
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1Note: $1/x^3=x^{-3}$ – J. W. Tanner Jan 22 '20 at 17:39
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1@kay ?? What division by 0? It is not true that $x^{\frac{1}{3}}=\frac{1}{x^3}$. Maybe you are thinking of $x^{-3}$ which not the cube root. – almagest Jan 22 '20 at 17:40
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1Why are you concerned about zero? We know that $0 \cdot 0 \cdot 0=0$, hence, by definition, $\sqrt[3] 0=0$ – Vasili Jan 22 '20 at 17:46
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1Absolutely right, I am sorry it was a moment of unclarity for me. – katrin Jan 22 '20 at 17:46
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$y=x^\frac13\implies x=y^3$, on the contrary, $y=\frac{1}{x^3}\implies x=(\frac 1y)^\frac13$
They aren't the same (except when $x=1$), because $\frac{1}{a^n}$ is written as $a^{-n}$, not $a^{\frac1n}$
Notice that $\frac{a^m}{a^n}=a^ma^{-n}=a^{m-n}$ holds with this notation
Rhys Hughes
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