Solve the following equation!

$$
\frac{(39-x) \sqrt[5]{x-6}-(x-6) \sqrt[5]{39-x}}{\sqrt[5]{39-x}-\sqrt[5]{x-6}}=30
$$
I. solution. Let's use the notations $u=\sqrt[5]{x-6}$ and $v=\sqrt[5]{39-x}$. The left side of the equation can then be written in the form

$$
\left(v^{5} u-u^{5} v\right) /(v-u)=u v\left(v^{4}-u^{4}\right) /(v-u)=u v(u+v)\left(u^{2}+v^{2}\right)
$$

Furthermore,

$$
u^{5}+v^{5}=(u+v)\left(u^{4}-u^{3} v+u^{2} v^{2}-u v^{3}+v^{4}\right)
$$

We get the following system of equations for $u$ and $v$:

$$
\begin{gathered}
u v(u+v)\left(u^{2}+v^{2}\right)=30 \\
(u+v)\left(u^{4}-u^{3} v+u^{2} v^{2}-u v^{3}+v^{4}\right)=33
\end{gathered}
$$

It is also useful to write the second factor on the left side of the second equation as a polynomial in $u^{2}+v^{2}$ and $u v$:

$$
u^{4}-u^{3} v+u^{2} v^{2}-u v^{3}+v^{4}=\left(u^{2}+v^{2}\right)^{2}-u^{2} v^{2}-u v\left(u^{2}+v^{2}\right)
$$

If we now divide the second equation by the first, we get the equation

$$
\frac{u^{2}+v^{2}}{u v}-\frac{u v}{u^{2}+v^{2}}-1=1.1
$$

(The right side of the first equation is positive, so we did not lose any roots in the division). For the variable $z=\frac{u^{2}+v^{2}}{u v}=\frac{u}{v}+\frac{v}{u}$, we thus get the equation $z-\frac{1}{z}=2.1$. This has two solutions: $z=2.5$ and $z=-0.4$. However, as the sum of a number and its reciprocal, $z$ cannot fall between -2 and 2, so only

$$
z=\frac{u}{v}+\frac{v}{u}=2.5 \quad \text { is possible. }
$$

This is a quadratic equation in $\frac{u}{v}$, which has two solutions: 2 and $\frac{1}{2}$. Therefore, the equation of the problem can only be satisfied by $x$ for which

$$
\frac{u}{v}=\sqrt[5]{\frac{x-6}{39-x}}=2 \quad \text { or } \quad \sqrt[5]{\frac{x-6}{39-x}}=\frac{1}{2}
$$

Solving these two linear equations, we get $x_{1}=38$ and $x_{2}=7$. By substitution, we can verify that both values are indeed solutions to the problem. II. solution. We again solve the system of equations (1) and (2). $u+v$ cannot be zero, since otherwise $u^{5}+v^{5}$ would also be zero. So let's multiply (1) by $\frac{11}{u+v}$, (2) by $\frac{10}{u+v}$, and subtract the first resulting equation from the second. Then we get the equation

$$
-11 u v\left(u^{2}+v^{2}\right)+10\left(u^{4}-u^{3} v+u^{2} v^{2}-u v^{3}+v^{4}\right)=0
$$

Rearranging the left side:

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

Here $v \neq 0$, since otherwise the left side of the problem's equation would also be zero. Therefore, we can divide by $v^{4}$, and for $\frac{u}{v}=t$ we get the

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

quartic, so-called reciprocal equation. We solve this in the usual way: first we divide by $t^{2}$, so the left side

$$
10\left(t^{2}+\frac{1}{t^{2}}\right)-21\left(t+\frac{1}{t}\right)+10=10\left(t+\frac{1}{t}\right)^{2}-21\left(t+\frac{1}{t}\right)-10
$$

can be written in the form, that is, for $t+\frac{1}{t}$ (again) the

$$
10\left(t+\frac{1}{t}\right)^{2}-21\left(t+\frac{1}{t}\right)-10=0
$$

quadratic equation.

$\left(t+\frac{1}{t}\right)$ is precisely the $z$ from the previous solution, so according to the completion of the first solution, $t+\frac{1}{t}=2.5$, from which the two solutions are 38 and 7.

Remark. Many reached the equation $u v(u+v)\left(u^{2}+v^{2}\right)=30$, but then they used the unique prime factorization, that is, they assumed that $u$ and $v$ are integers. However, the problem was to find all real solutions. (It is another matter that it turns out during the solution that $u$ and $v$ are integers.)