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How can I show $F:\mathbb{R}^2\rightarrow \mathbb{R}^2$ given by $F(x,y)=(x^3-3xy^2, 3x^2y-y^3)$ is surjective? I've tried solving for $(u,v)=(x^3-3xy^2, 3x^2y-y^3)$ (not solving explicitly, just trying to show solutions exist) but I end up with some square roots that I don't know are real.

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HINT: Consider $z \rightarrow z^3$ where $z \in \mathbb{C}$. Consider any $(a,b) \in \mathbb{R}^2$ and look at $\sqrt[3]{a+ib}$. – user17762 Nov 9 '11 at 21:15
Maybe you could check that, e.g., the two functions are neither linearly-dependent from each other, and that they are not both in the same (algebraic) variety. – gary Nov 9 '11 at 21:17
A slight variant of Sivaram: use polar coordinates: $(re^{i\t})^{1/3}=r^{1/3}e^{i2k\pi/3}$ ; select just one of the three roots. – gary Nov 9 '11 at 21:24

Rewriting in polar (suggested by gary) or complex (suggested by user177622) form is the way to go. I'll use the polar coordinates $(r,\theta)$. Observing that $F: (r,\theta)\mapsto (r^3,3\theta)$, one easily concludes that the image of the circle of radius $r$ under $F$ is the circle of radius $r^3$. Since $r\mapsto r^3$ is a surjection of $[0,\infty)$ onto itself, $F$ is surjective.

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