5. The area of the figure enclosed by the two curves $y=x^{3}$ and $x=y^{3}$ on the Cartesian plane is $\qquad$ .
5.1.

Since the two curves are symmetric with respect to the origin, it is only necessary to calculate the area $A$ of the figure enclosed by the two curves in the first quadrant.
When $x>1$, $x^{3}>\sqrt[3]{x}$;
When $0<x<1$, $x^{3}<\sqrt[3]{x}$.
Therefore, the two curves have a unique intersection point $(1,1)$ in the first quadrant.
$$
\begin{array}{l}
\text { Also, } A=\int_{0}^{1}\left(\sqrt[3]{x}-x^{3}\right) \mathrm{d} x \\
=\left.\left(\frac{3}{4} x^{\frac{4}{3}}-\frac{x^{4}}{4}\right)\right|_{0} ^{1}=\frac{3}{4}-\frac{1}{4}=\frac{1}{2},
\end{array}
$$

Therefore, the area of the figure enclosed by the two curves is $2 A=1$.