Solve the following equation:

$$
\sqrt[3]{\frac{a+x}{2 a-x}}+\sqrt[3]{\frac{a-x}{2 a+x}}=1
$$
Let's raise both sides of the equation to the third power. The cube of the left side, according to the identity \((A+B)^{3}=A^{3}+B^{3}+3 A B(A+B)\), can be immediately written in place of the right side parenthesis:

\[
\frac{a+x}{2 a-x}+\frac{a-x}{2 a+x}+3 \sqrt[3]{\frac{a^{2}-x^{2}}{4 a^{2}-x^{2}}}=1
\]

By rearranging, raising to the third power again, and rearranging:

\[
\begin{gathered}
\sqrt[3]{\frac{a^{2}-x^{2}}{4 a^{2}-x^{2}}}=\frac{-x^{2}}{4 a^{2}-x^{2}} \\
a^{2}\left(4 a^{2}-3 x^{2}\right)^{2}=0
\end{gathered}
\]

It cannot be \(a=0\), otherwise equation (1) has no solution because its left side is -2 for all \(x\) values different from 0. Therefore, only such \(x\) numbers can be roots for which the expression in parentheses in (2) is 0. These two numbers are:

\[
x_{1}=2 a / \sqrt{3}, \quad x_{2}=-x_{1}
\]

Equation (2) was not formed from (1) by purely equivalent transformations, so we decide by substitution whether \(x_{1}\) is a root of (1). Substituting this into the expressions under the cube roots:

\[
\begin{gathered}
\frac{a(1+2 / \sqrt{3})}{a(2-2 / \sqrt{3})}=\frac{\sqrt{3}+2}{2(\sqrt{3}-1)}=\frac{2(\sqrt{3}+4)}{4(\sqrt{3}-1)}=\frac{(\sqrt{3}+1)^{2}}{4(\sqrt{3}-1)}=\frac{(\sqrt{3}+1)^{3}}{8} \\
\frac{a(1-2 / \sqrt{3})}{a(2+2 / \sqrt{3})}=\left(\frac{1-\sqrt{3}}{2}\right)^{3}
\end{gathered}
\]

Thus, the left side of (1) is 1, so \(x_{1}\) is a root of the given equation. Similarly, \(x_{2}\) is also a root, because substituting \(-x\) for \(x\) the two expressions under the cube roots transform into each other.