Showing a Variety is Rational? 
I'm trying to show that the following varieties are rational: $V_1=V(y^2z-x^3)$ and $V_2=V(xyz-x^3-y^3)$. 

But I can't think of how to show they are birationally equivalent to $\mathbb{A}^n$ or $\mathbb{P}^n$ for some $n$. I've tried parametrizing the variables with the relations given, but that got me nowhere. Thank you.
 A: For the variety $V_1$  described $zy^2=x^3$, concentrate on the open subset of $V_1$ where $y\ne0$. There we can write $z=  x^3/y^2$. This open set is  the image of the morphism from $\mathbf{A}^2\to\mathbf{A}^3$  given by $(s,t)\mapsto (s,t, s^3/t^2)$ providing a rational parametrisation for $V_1$.
And for $V_2$ a suitable open set is the image of $\displaystyle (s,t)\mapsto \bigg(s,t, \frac{s^3+t^3}{st}\bigg)$.
By proof analysis we can see that an algebraic  variety defined by a single equation is rational if it is of degree 1 in one of the variables. The equation defining the variety is of the form $yf(x_1,x_2,\ldots,x_n)+c =0$.Then the image of the  map from $n$-dimensional affine space defined by $(t_1,t_2,\ldots,t_n)\mapsto\big (t_1,t_2,\ldots,t_n, -c/f(t_1,t_2,\ldots,t_n)\,\big )$ is the opne set of the variety that is in bijective correspondence with an open set of the affine space (where $f\ne0)$
A: Hint: In the affine open set where $z \neq 0$, we can dehomogenize the above equations by setting $z = 1$.  In this way, we obtain $V_1: Y^2 = X^3$, the cuspidal cubic and $V_2: X^3 + Y^3 - XY = 0$, the folium of Descartes.  Both have well-known parametrizations that can be obtained by considering a pencil of lines through the singular point at the origin.  
