6.040. $\sqrt[3]{24+\sqrt{x}}-\sqrt[3]{5+\sqrt{x}}=1$.

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6.040. $\sqrt[3]{24+\sqrt{x}}-\sqrt[3]{5+\sqrt{x}}=1$.
## Solution.

Domain of definition: $x \geq 0$.

Raising both sides of the equation to the third power, we get

$$
\begin{aligned}
& 24+\sqrt{x}-3 \sqrt[3]{(24+\sqrt{x})^{2}(5+\sqrt{x})}+3 \sqrt[3]{(24+\sqrt{x})(5+\sqrt{x})^{2}}-5-\sqrt{x}=1 \Leftrightarrow \\
& \Leftrightarrow-3 \sqrt[3]{(24+\sqrt{x})}(5+\sqrt{x})(\sqrt[3]{24+\sqrt{x}}-\sqrt[3]{5+\sqrt{x}})=-18
\end{aligned}
$$

Since $\sqrt[3]{24+\sqrt{x}}-\sqrt[3]{5+\sqrt{x}}=1$ by the condition, we obtain

$\sqrt[3]{(24+\sqrt{x})}(5+\sqrt{x})=6 \Leftrightarrow(24+\sqrt{x})(5+\sqrt{x})=216$,

$(\sqrt{x})^{2}+29 \sqrt{x}-96=0$.

From which $\sqrt{x}=3, \sqrt{x}=-3$ (not suitable). Therefore, $x=9$.

Answer: $x=9$.