Solve the following equation:

$$
\sqrt[5]{x+1}-\sqrt[5]{x-1}=\sqrt[5]{2}
$$
Divide the equation by $\sqrt[5]{2}$.

# Then

$$
\sqrt[5]{\frac{x+1}{2}}-\sqrt[5]{\frac{x-1}{2}}=1
$$

Introduce new variables, thereby transforming the irrational equation into a system of rational equations:

$$
\sqrt[5]{\frac{x+1}{2}}=u \quad \sqrt[5]{\frac{x-1}{2}}=v
$$

According to the condition, $u-v=1$ and since the difference of the bases is 1, then $u^{5}-v^{5}=1$. Raising both sides of the first equation to the fifth power, the resulting $(u-v)^{5}=1$ equation's left side can be expanded in terms of powers of $u$ and $v$:

$$
u^{5}-5 u^{4} v+10 u^{3} v^{2}-10 u^{2} v^{3}+5 u v^{4}-v^{5}=1
$$

Using the second equation, we get

$$
\begin{aligned}
& -5 u v\left(u^{3}-2 u^{2} v+2 u v^{2}-v^{3}\right)=0 \\
& -5 u v(u-v)\left(u^{2}-u v+v^{2}\right)=0
\end{aligned}
$$

Since according to the condition $u-v \neq 0$, this product is zero when

$$
\begin{array}{cl}
\operatorname{or} & u=0, \quad \text { i.e. } \quad x=-1 \\
\text { or } & v=0, \quad \text { i.e. } \quad x=1
\end{array}
$$

or using that according to our first equation $u=1+v$; if

$$
u^{2}-u v+v^{2}=(1+v)^{2}-(1+v) v+v^{2}=1+v+v^{2}=0
$$

The solution to the latter equation is

$$
v=-\frac{1}{2} \pm \frac{i \sqrt{3}}{2}=\frac{x-1}{2}
$$

From this,

$$
x= \pm i \sqrt{3}
$$

Considering that the root extraction is multivalued in the complex numbers, taking appropriate values of the fifth roots, this value of $x$ can also be a solution.