Answer

Always solve such questions using the cumulative distribution approach.

\[ \begin{alignat}{2} P(X \leq x) &= \begin{cases} 0 &\mbox{$x < 0$}\newline \frac{1}{2} x &\mbox{$0 \leq x \leq 2$}\newline 1 &\mbox{$2 < x$} \end{cases}\newline P(Y \leq y) &= P(X^{3} \leq y) = P(X \leq y^{\frac{1}{3}})\newline &= \begin{cases} 0 &\mbox{$y < 0$}\newline \frac{1}{2} y^{\frac{1}{3}} &\mbox{$0 \leq y^{\frac{1}{3}} \leq 2$}\newline 1 &\mbox{$2 < y^{\frac{1}{3}}$} \end{cases}\newline f_{Y}(y) &= \frac{dP(Y <= y)}{dy}(y)\newline &= \begin{cases} 0 &\mbox{$y < 0$}\newline \frac{1}{6} y^{\frac{-2}{3}} &\mbox{$0 \leq y \leq 8$}\newline 0 &\mbox{$8 < y$} \end{cases} \end{alignat} \]