How is the Dirac delta $\delta (x^3)$ different from $\delta(x)$?

Question

How is Dirac delta $\delta (x^3 )$ different from $\delta (x)$? It is my understanding that $$ \delta [\psi]:=\langle\delta,\psi\rangle=\int_{\Omega}\delta(x)\psi(x)\,dx= \begin{cases} \psi(0) & \text{if $0\in \Omega$}\\ 0 & \text{otherwise} \end{cases} $$ But $x^3=0$ iff $x=0$ so does this not imply, $$\delta(x^3) = \delta(x) = 0$$ Apologies if I am completely misunderstanding the topic, I am very new to distributions.

Answer 1

Let $\psi$ be a test function, i.e. $\psi \in \mathcal{C}^{\infty} _{c}(\mathbb{R})$, then: $$ \int \delta(x^3)\psi (x) dx = \int \delta(y) \psi (y^{\frac{1}{3}}) \frac{1}{3y^{\frac{2}{3}}} dy $$ Note that our new test function $\psi (y^{\frac{1}{3}}) \frac{1}{3y^{2/3}}$ is not smooth, hence we cannot define $\delta(x^3)$ as a distribution. In general, if $u$ is a distribution, then $u \circ g$ is a distribution as well if both $g$ and its inverse are smooth. In such a case: $\langle u \circ g, \psi \rangle := \langle u, \psi \circ g^{-1} |\text{det } dg^{-1}| \rangle$. Please note that the integration above is an abuse of notation.