Determine all real numbers $x$ for which $x^{3 / 5}-4=32-x^{2 / 5}$.
Let $t=x^{1 / 5}$. Thus, $x^{2 / 5}=t^{2}$ and $x^{3 / 5}=t^{3}$.

Therefore, the following equations are equivalent:

$$
\begin{aligned}
x^{3 / 5}-4 & =32-x^{2 / 5} \\
t^{3}+t^{2}-36 & =0 \\
(t-3)\left(t^{2}+4 t+12\right) & =0
\end{aligned}
$$

Thus, $t=3$ or $t^{2}+4 t+12=0$.

The discriminant of this quadratic equation is $4^{2}-4(1)(12)<0$, which means that there are no real values of $t$ that are solutions. This in turn means that there are no corresponding real values of $x$.

This gives $x^{1 / 5}=t=3$ and so $x=3^{5}=243$ is the only solution.

ANSWER: $3^{5}=243$