6.037. $\sqrt[3]{1+\sqrt{x}}+\sqrt[3]{1-\sqrt{x}}=2$.
Solution.

Domain of definition: $x \geq 0$.

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

$$
\begin{aligned}
& 1+\sqrt{x}+3 \sqrt[3]{(1+\sqrt{x})^{2}(1-\sqrt{x})}+3 \sqrt[3]{(1+\sqrt{x})(1-\sqrt{x})^{2}}+1-\sqrt{x}=8 \Leftrightarrow \\
& \Leftrightarrow 3 \sqrt[3]{(1+\sqrt{x})^{2}(1-\sqrt{x})}+3 \sqrt[3]{(1+\sqrt{x})(1-\sqrt{x})^{2}}=6 \Leftrightarrow \\
& \Leftrightarrow 3 \sqrt[3]{(1+\sqrt{x})(1-\sqrt{x})}(\sqrt[3]{1+\sqrt{x}}+\sqrt[3]{1-\sqrt{x}})=6
\end{aligned}
$$

Since $\sqrt[3]{1+\sqrt{x}}+\sqrt[3]{1-\sqrt{x}}=2$, the equation becomes:

$3 \sqrt[3]{(1+\sqrt{x})(1-\sqrt{x})} \cdot 2=6 \Leftrightarrow \sqrt[3]{(1+\sqrt{x})(1-\sqrt{x})}=1 \Leftrightarrow \sqrt[3]{1-x}=1 \Leftrightarrow$ $\Leftrightarrow 1-x=1, x=0$.

Answer: $x=0$.