6.054. $\sqrt{5+\sqrt[3]{x}}+\sqrt{5-\sqrt[3]{x}}=\sqrt[3]{x}$.
Solution.

Domain of definition: $\left\{\begin{array}{l}5+\sqrt[3]{x} \geq 0 \\ 5-\sqrt[3]{x} \geq 0\end{array} \Leftrightarrow -125 \leq x \leq 125\right.$.

By squaring both sides of the equation, we get the equation

$$
\begin{aligned}
& 5+\sqrt[3]{x}+2 \sqrt{(5+\sqrt[3]{x})(5-\sqrt[3]{x})}+5-\sqrt[3]{x}=\sqrt[3]{x^{2}} \Leftrightarrow \\
& \Leftrightarrow 2 \sqrt{25-\sqrt[3]{x^{2}}}=\sqrt[3]{x^{2}}-10 \Rightarrow 100-4 \sqrt[3]{x^{2}}=\sqrt[3]{x^{4}}-20 \sqrt[3]{x^{2}}+100 \Leftrightarrow \\
& \Leftrightarrow \sqrt[3]{x^{4}}-16 \sqrt[3]{x^{2}}=0 \Leftrightarrow \sqrt[3]{x^{2}}\left(\sqrt[3]{x^{2}}-16\right)=0
\end{aligned}
$$

from which $\sqrt[3]{x^{2}}=0, x_{1}=0$, or $\sqrt[3]{x^{2}}-16=0, \sqrt[3]{x^{2}}=16, x_{2}=64$. Upon verification, $x_{1}=0$ does not satisfy the original equation.

Answer: $x=64$.