Solve the following equation:

$$
\frac{(39-x) \sqrt[5]{x-6}-(x-6) \sqrt[5]{39-x}}{\sqrt[5]{39-x}-\sqrt[5]{x-6}}=30
$$
Solution. Let $\sqrt[5]{39-x}=a \Rightarrow 39-x=a^{5}, \sqrt[5]{x-6}=b \Rightarrow x-6=b^{5}$. Then the original equation is:

$$
\frac{a^{5} b-a b^{5}}{a-b}=30
$$

At the same time, it also holds that $a^{5}+b^{5}=33$. Since $a-b \neq 0$ (the denominator cannot be zero), the left side of the equation can be simplified by canceling $(a-b)$:

$$
\frac{a^{5} b-a b^{5}}{a-b}=\frac{a b\left(a^{4}-b^{4}\right)}{a-b}=\frac{a b(a-b)(a+b)\left(a^{2}+b^{2}\right)}{a-b}=a b(a+b)\left(a^{2}+b^{2}\right)=30
$$

Dividing this by the equation $a^{5}+b^{5}=33$ obtained from the condition (none of the factors can be zero, as the right side of the equations is non-zero), we get:

$$
\frac{a b(a+b)\left(a^{2}+b^{2}\right)}{(a+b)\left(a^{4}-a^{3} b+a^{2} b^{2}-a b^{3}+b^{4}\right)}=\frac{30}{33}=\frac{10}{11}
$$

After simplification, multiplication, and rearrangement, we obtain a homogeneous quartic equation with symmetrically placed coefficients:

$$
10 a^{4}-21 a^{3} b+10 a^{2} b^{2}-21 a b^{3}+10 b^{4}=0
$$

Since $b$ is not zero, we can divide the equation by $b^{4}$. Then we introduce a new unknown $k$ for $\frac{a}{b}$:

$$
10 k^{4}-21 k^{3}+10 k^{2}-21 k+10=0
$$

This is a so-called reciprocal equation, which can always be reduced to a quadratic equation. Since $k=0$ is not a solution to the equation, we can divide by $k^{2}$ and rearrange the terms:

$$
\begin{array}{r}
10 k^{2}+\frac{10}{k^{2}}-21 k-\frac{21}{k}+10=0 \\
10\left(k^{2}+\frac{1}{k^{2}}\right)-21\left(k+\frac{1}{k}\right)+10=0 \\
10\left(k+\frac{1}{k}\right)^{2}-21\left(k+\frac{1}{k}\right)-10=0
\end{array}
$$

Now, with the substitution $u=k+\frac{1}{k}$, we arrive at a quadratic equation:

$$
10 u^{2}-21 u-10=0
$$

The solutions to this equation are $u=\frac{5}{2}$ and $u=-\frac{2}{5}$. The latter does not yield a real solution because $k+\frac{1}{k}$ for real numbers is at least 2 in absolute value. From the former:

$$
k+\frac{1}{k}=\frac{5}{2}, \quad \text { i.e., } \quad k=2, \quad k=\frac{1}{2}
$$

From here, the solutions to the original equation are:

$$
\begin{aligned}
& \frac{\sqrt[5]{39-x}}{\sqrt[5]{x-6}}=2 \Longrightarrow \frac{39-x}{x-6}=32 \Longrightarrow x=7 \\
& \frac{\sqrt[5]{39-x}}{\sqrt[5]{x-6}}=\frac{1}{2} \Longrightarrow \frac{39-x}{x-6}=\frac{1}{32} \Longrightarrow x=38
\end{aligned}
$$

Substitution confirms that both roots satisfy the equation.