Note that we can re-write the given equation as $$\sqrt[3]{x-\frac{3}{8}}=x^{3}+\frac{3}{8}$$ Furthermore, the functions of $x$ on either side, we see, are inverses of each other and increasing. Let $f(x)=\sqrt[3]{x-\frac{3}{8}}$. Suppose that $f(x)=y=f^{-1}(x)$. Then, $f(y)=x$. However, if $x<y$, we have $f(x)>f(y)$, contradicting the fact that $f$ is increasing, and similarly, if $y<x$, we have $f(x)<f(y)$, again a contradiction. Therefore, if $f(x)=f^{-1}(x)$ and both are increasing functions in $x$, we require $f(x)=x$. This gives the cubic $$x^{3}-x+\frac{3}{8}=0 \rightarrow\left(x-\frac{1}{2}\right)\left(x^{2}+\frac{1}{2} x-\frac{3}{4}\right)=0$$ giving $x=\frac{1}{2}, \frac{-1 \pm \sqrt{13}}{4}$.