<s>[INST] Calculate the probability density function of two independent random variables: (R^3 + r^3)^(1/3) Random variables R and r are independent, both of them are uniform distributed and greater than zero. R distributes in (a,b), r in (c,d). I tried to solve the probability density function of (R^3 + r^3)^(1/3), but I couldn't figure it out. so any one can help me out?
Question Update: c > a > 0, a + t = b , c + t = d, t > 0, c - a => t
 [/INST] I recommend doing this one step at a time.
First let's find the CDFs and densities of $R^3$ and $r^3$:
$$F_{R^3}(x)=P(R^3 \leq x) = P(R \leq x^{1/3}) = \int_{-\infty}^{x^{1/3}}\frac{1}{b-a}1_{a \leq t^{1/3} \leq b} \ dt = \begin{cases}0 & x < a^3 \\ \frac{x^{1/3}-a}{b-a}  & a^3\leq x\leq b^3 \\ 1 & x>b^3\end{cases}$$
so 
$$f_{R^3}(x) = \frac{d}{dx}F_{R^3}(x) = \begin{cases} 0 & x<a^3, x>b^3 \\ \frac{1}{3(b-a)}x^{-2/3}& a^3 \leq x \leq b^3 \end{cases}.$$
Similarly you can find $F_{r^3}(x)$ and $f_{r^3}(x)$. 
Then you can find the density of the sum $R^3+r^3$ using the fact that $R^3$ and $r^3$ are independent:
$$f_{R^3+r^3}(x) = \int_{-\infty}^{\infty} f_{R^3}(x-t)f_{r^3}(t) \ dt.$$
From that you can find $F_{R^3+r^3}(x) = \int_{-\infty}^x f_{R^3+r^3}(t) \ dt$. Finally, from that you can find 
$$F_{(R^3+r^3)^{1/3}}(x) = P((R^3+r^3)^{1/3} \leq x) = P(R^3+r^3 \leq x^3)=F_{R^3+r^3}(x^3)$$
and 
$$f_{(R^3+r^3)^{1/3}}(x) = \frac{d}{dx}F_{(R^3+r^3)^{1/3}}(x).$$
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