<s>[INST] Probability distribution of a uniform distribution to the third power I have to find a explicit form of probability distribution of $X^3$, if $X \ \mathtt{\sim} \ U[a, b], \ -\infty < a < b < \infty$.
So far I've succesfully done a simpler version, when it's $\ U[0, a], \ \ 0 < a < \infty$:
$F_{X^3}(x) = \mathbb{P}(X^3 \leq x) = \mathbb{P}(X \in [0, \sqrt[3]{x}]) = \int^{\sqrt[3]{x}}_0 \frac{dx}{a} = \frac{\sqrt[3]{x}}{a},$
$f_{X^3}(x) = \frac{dF_{X^3}(x)}{dx} = \dfrac{1}{3ax^{\frac{2}{3}}}$,
where the support is $0 \leq x \leq a^3$.
However, the way of thinking that is above got me lost at the full version of the uniform distribution.
$F_{X^3}(x) = \mathbb{P}(X^3 \leq x) = \mathbb{P}(X \in [-\sqrt[3]{x}, \sqrt[3]{x}]) = \int^{\sqrt[3]{x}}_{-\sqrt[3]{x}} \frac{dx}{b-a} = \frac{2\sqrt[3]{x}}{b-a}$
$f_{X^3}(x) = \frac{dF_{X^3}(x)}{dx} = \dfrac{2}{3x^{\frac{2}{3}} \cdot (b-a)}$, where the support is...?
If the support would be $a^3 \leq x \leq b^3$ and $a < 0$ then the integral
$\int^{b^3}_{a^3} f(x)$ would have no sense for negative values, whereas it should equal $1$. 
Or where else did I made a mistake?
 [/INST] Tip: Don't use $x$ both in the domain's bounds and as the integration's variable.   That inevitably leads to confusion.   Additionally, we instinctively associate lower and upper case letters as referring to the same thing, so that's the best one to change.
Now if the support for $X$ is $a\leq X\leq b$, then the support for $X^3$ is $a^3\leq X^3\leq b^3$.   There are no folds, because $x\mapsto x^3$ is an increasing function over the support for $X$.
$$\begin{align}F_{X^3}(y)&= \mathsf P(X\leq y^{1/3})\\[1ex] &=\mathbf 1_{a^3\leq y\leq b^3} \int_a^{y^{1/3}}\tfrac 1{b-a} \mathrm d x+\mathbf 1_{y>b^3}\\[1ex]&= \phantom{\tfrac{y^{1/3}-a}{b-a}\mathbf 1_{a^3\leq y\leq b^3}+\mathbf 1_{b^3<y}}\\[3ex] f_{X^3}(y)&=\left\lvert\dfrac{\mathrm d ~~}{\mathrm d y}F_{X^3}(y)\right\rvert\\[1ex]&=\phantom{\tfrac{\lvert y^{-2/3}\rvert}{b-a}\mathbf 1_{a^3\leq y\leq b^3}}\end{align}$$
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