= Solution
The parametrization $t\mapsto(t^3,t^2)$ maps $\mathbb A^1$ to the cusp. Off the origin its inverse is $t=x/y$, since $(x/y)^2=y$ and $(x/y)^3=x$ on $x^2=y^3$ with $y\ne0$. This is an isomorphism between dense open subsets, so \b[the cusp is birational to $\mathbb A^1$].
At the origin the differential of the defining polynomial is zero, giving a two-dimensional <Zariski tangent space>. The tangent space of $\mathbb A^1$ at every point is one-dimensional. An isomorphism induces tangent-space isomorphisms, so \b[the cusp is not isomorphic to $\mathbb A^1$]. The rational inverse cannot extend regularly through the singular point.
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