Normalization of y squared equals x times x minus one squared ($y^2=x(x-1{)}^2$)

For the nodal affine cubic $y^2=x(x-1{)}^2$, the rational parametergives $x=t^2$ and $y=t(t^2-1)$. Its function field is $k(t)$, so its smooth projective normalization is ${\mathbb{P}}^1$ and has genus zero.