Show that the algebraic curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=1$ can be given by a polynomial of degree $6$? I imagine that  this should be done in the following way: There is a polynomial $P$ such that:
$$P(x^{\frac{2}{3}}+y^{\frac{2}{3}})=P$$
My first guess (obvious observation?) is that it can't be a polynomial in one variable, otherwise one of the variables will have a rational power. So the number of variables must be $2$. I've tried this: 
$$(x^n+y^m)(x^{\frac{2}{3}}+y^{\frac{2}{3}})=(x^n+y^m)$$
$$(x^n+y^m)x^{\frac{2}{3}}+(x^n+y^m)y^{\frac{2}{3}}=x^n+y^m$$
$$x^nx^{\frac{2}{3}}+y^mx^{\frac{2}{3}}+x^ny^{\frac{2}{3}}+y^my^{\frac{2}{3}}-x^n-y^m=0$$
$$x^nx^{\frac{2}{3}}+\overbrace{y^m(x^{\frac{2}{3}}-1)+x^n(y^{\frac{2}{3}}-1)}^{\text{Annoying cross-terms }}+y^my^{\frac{2}{3}}=0$$
So this seems to show that we need to have at least one of the sums $n+2/3,m+2/3$ to be equal to $6$. But what should be the value of the other? And how do we eliminate the annoying cross-terms?
I understand also that $P$ could be $x^{a_1}+y^{b_1}+x^{a_2}+y^{b_2}+\dots +x^{a_t}+y^{b_t}$ for a certain $t$ but I don't know how to look into the entire collection of P's to find for it and also I don't know how to cut the general annoying cross-terms that this would give me. If possible, I'd like to have only some hints.

Hint:
$$
\begin{align}
1=\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}\right)^3 & = \left(\sqrt[3]{x^2}\right)^3 + 3\;\sqrt[3]{x^2}\;\sqrt[3]{y^2}\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}\right)+\left(\sqrt[3]{y^2}\right)^3 \\
 & =x^2 + y^2 + 3\;\sqrt[3]{x^2\,y^2}
\end{align}
$$
Regroup, and raise to the $3^{rd}$ power one more time.
