II. (40 points) Find the real solution of the equation
$$
\sqrt[3]{x(3+\sqrt{8 x-3})-1}+\sqrt[3]{x(3-\sqrt{8 x-3})-1}=1
$$
$$
\begin{array}{l}
\text { II. Let } A=\sqrt[3]{x(3+\sqrt{8 x-3})-1}, \\
B=\sqrt[3]{x(3-\sqrt{8 x-3})-1} . \\
\text { Then } A^{3}+B^{3}=6 x-2, A B=1-2 x .
\end{array}
$$
$$
\text { Let } t=A+B \text {. }
$$

Notice that,
$$
(A+B)^{3}=A^{3}+B^{3}+3 A B(A+B) \text {. }
$$

Then $t^{3}-3(1-2 x) t-(6 x-2)=0$, which is
$$
\begin{array}{l}
(t-1)\left(t^{2}+t+6 x-2\right)=0 . \\
\text { Also, } t^{2}+t+6 x-2=\left(t+\frac{1}{2}\right)^{2}+6\left(x-\frac{3}{8}\right),
\end{array}
$$

When $x=\frac{3}{8}$, $t=A+B=2 \sqrt[3]{3 x-1}=1$.
Thus, $t^{2}+t+6 x-2>0$.
Therefore, for any $x \geqslant \frac{3}{8}$, we have
$$
t^{2}+t+6 x-2>0 \text {. }
$$

Hence, $t=1$.
In conclusion, the set of real solutions to the original equation is
$$
\left\{x \left\lvert\, x \geqslant \frac{3}{8}\right.\right\} .
$$