6.181. $\sqrt[3]{x+a}+\sqrt[3]{x+a+1}+\sqrt[3]{x+a+2}=0$.
## Solution.

Rewriting the equation as $\sqrt[3]{x+a}+\sqrt[3]{x+a+1}=-\sqrt[3]{x+a+2}$ and cubing both sides, we get

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

Since $\sqrt[3]{x+a}+\sqrt[3]{x+a+1}=-\sqrt[3]{x+a+2}$, we have

$$
\begin{aligned}
& -\sqrt[3]{(x+a)(x+a+1)(x+a+2)}=-(x+a+1) \Leftrightarrow \\
& \Leftrightarrow \sqrt[3]{(x+a)(x+a+1)(x+a+2)}=(x+a+1) \Leftrightarrow \\
& \Leftrightarrow(x+a)(x+a+1)(x+a+2)=(x+a+1)^{3} \Leftrightarrow \\
& \Leftrightarrow(x+a+1)\left((x+a)(x+a+2)-(x+a+1)^{2}\right)=0 \Leftrightarrow \\
& (x+a+1)(-1)=0, \text { from which } x=-(a+1) .
\end{aligned}
$$

Answer: $x=-(a+1)$.