6.158. $\sqrt[3]{9-\sqrt{x+1}}+\sqrt[3]{7+\sqrt{x+1}}=4$.
## Solution.

Domain of definition: $x+1 \geq 0$ or $x \geq-1$.

Let $\sqrt{x+1}=y \geq 0$. The equation in terms of $y$ becomes $\sqrt[3]{9-y}+\sqrt[3]{7+y}=4$. Raising both sides of the equation to the third power, we get

$$
\begin{aligned}
& 9-y+\sqrt[3]{(9-y)^{2}(7+y)}+3 \sqrt[3]{(9-y)(7+y)^{2}}+7+y=64 \Leftrightarrow \\
& \Leftrightarrow 3 \sqrt[3]{(9-y)^{2}(7+y)}+3 \sqrt[3]{(9-y)(7+y)^{2}}=48 \Leftrightarrow \\
& \Leftrightarrow 3 \sqrt[3]{(9-y)(7+y)} \times(\sqrt[3]{9-y}+\sqrt[3]{7+y})=48
\end{aligned}
$$

Since by the condition $\sqrt[3]{9-y}+\sqrt[3]{7+y}=4$, then

$$
12 \sqrt[3]{(9-y)(7+y)}=48 \Leftrightarrow(9-y)(7+y)=64, y^{2}-2 y+1=0,(y-1)^{2}=0
$$

from which $y_{1,2}=1$.

Then $\sqrt{x+1}=1, x=0$.

Answer: $x=0$.