The following equation graphs a love heart,
$x^2+(y- \sqrt[3]{x^2})^2=1$
And since (please correct me if I'm wrong)
$\sqrt[3]{x^2}=x^{2/3}$
I should be able to write it like this:
$x^2+(y- x^{2/3})^2=1$
However when I graph the latter I only get the positive side of the heart, not the negative side. Can someone tell me why this is?
I think the graphing software is just wrong in this case.
Consider $x=-1.$
Then $x^2 = 1$, so $(x^{2})^{1/3} = 1$.
Also $x^{1/3} = -1$, so $(x^{1/3})^2 = 1.$
Hence $(x^{2})^{1/3} = (x^{1/3})^2$ for $x$ equal to $-1$.
This suggests to me that $\sqrt[3]{x^2}=x^{2/3}$ holds for all real $x$, not just the positive ones.
Edit. One possibility is that the moment you raise $x$ to the power of a non-integer, the graphing software automatically interprets $x^r$ as $\exp(r \log x),$ which only makes sense if $x$ is positive. According to this convention, $x^{2/3}$ is defined only for positive $x$, while $\sqrt[3]{x^2}$ is defined for all $x$.