Solve the following equation:

$$
\left(11 x^{3}+80\right) \sqrt[3]{x+11}+\left(x^{3}-92\right) \sqrt[3]{92 x-80}=0
$$
Solution. This equation cannot have a solution $x \geq \sqrt[3]{92}$, because then both products would be non-negative, and at least one would be positive. It is also easily seen that $\sqrt[3]{x+11} \neq 0$ and $x^{3}-92 \neq 0$, so we can divide by these:

$$
\frac{11 x^{3}+80}{92-x^{3}}=\frac{\sqrt[3]{92 x-80}}{\sqrt[3]{x+11}}
$$

Let

$$
y=\frac{\sqrt[3]{92 x-80}}{\sqrt[3]{x+11}}, \quad \text { then } \quad y^{3}=\frac{92 x-80}{x+11}, \quad \text { from which } \quad x=\frac{11 y^{3}+80}{92-y^{3}}
$$

Thus, the function

$$
g: x \mapsto \frac{\sqrt[3]{92 x-80}}{\sqrt[3]{x+11}}
$$

is the inverse of the function

$$
f: x \mapsto \frac{11 x^{3}+80}{92-x^{3}}
$$

The function $f(x)$ is strictly increasing on the interval $(-\infty, \sqrt[3]{92})$, since there the numerator is strictly increasing, and the denominator is positive and strictly decreasing. It is also easily verifiable that the range of $f$ defined on $(-\infty, \sqrt[3]{92})$ is the set of real numbers. From this it follows that the range of the function $g$, which is defined and strictly increasing on the entire set of real numbers, is $(-\infty, \sqrt[3]{92})$. If $a$ is a solution to our equation, that is, $f(a)=g(a)=b$ holds for some $b$, then in the case of $a<b$ we have $b=f(a)>f(b)=f(g(a))=a$; both lead to a contradiction. Therefore, the equation has a solution $a$ if and only if $f(a)=\sqrt{1}$, that is,

$$
\begin{aligned}
\frac{11 a^{3}+80}{92-a^{3}} & =a \\
a^{4}+11 a^{3}-92 a+80 & =0
\end{aligned}
$$

The left side can be factored as:

$$
\left(a^{2}-3 a+2\right)\left(a^{2}+14 a+40\right)=0
$$

Thus, the solutions to the equation are $-10, -4, 1$, and $2$. [^0]


[^0]:    ${ }^{1}$ See Ábrahám Gábor: On equations of the type $A z f^{-1}(x)=f(x)$, or the responsibility of writers and other interesting things, KöMaL (2010/9), 514-527. о.