Fulton Surjective Polynomial Map Chapter 2 I am doing exercices from Fulton's "Algebraic Curves" Chapter 2, it deals with polynomial maps. In exercice 13, we have to show that : 
$  \begin{equation*}
\begin{array}{ccccc}
f & : & \mathbb{A}^1(\mathbb{K}) & \to & V(X^2-Y^3,Y^2-Z^3) \subset \mathbb{A}^3(\mathbb{K})  \\
 & &t& \mapsto & (t^9,t^6,t^4)  \\
\end{array}
\end{equation*}
$
is a surjective map. In other exercices, I prove the surjectivity with the "construction of the map". For exemple, for the singular cubic $V(Y^2-X^3)$, I draw a line from $(0,0)$ to another point $(x,y) \in V(Y^2-X^3)$ and we get a polynomial map. But in our case, I don't know how to do...

You need to show that each solution of $X^2-Y^3=Y^2-Z^3$
has the form $(X,Y,Z)=(t^9,t^6,t^4)$. Let's assume $XYZ\ne0$. Let
$t=XY^{-2}Z$. Then $t^9=X^9Y^{-18}Z^9=X(X^2Y^{-3})^4(Y^{-2}Z^3)^3=X$
etc.
