$6.294 \sqrt[3]{(2-x)^{2}}+\sqrt[3]{(7+x)^{2}}-\sqrt[3]{(7+x)(2-x)}=3$.
Solution. Let $u=\sqrt[3]{2-x}, v=\sqrt[3]{7+x}$, then $\left\{\begin{array}{l}u^{2}+v^{2}-u v \doteq 3, \\ u^{3}+v^{3}=9\end{array} \Leftrightarrow\right.$

$$
\begin{aligned}
& \Leftrightarrow\left\{\begin{array} { l } 
{ u ^ { 2 } + v ^ { 2 } - u v = 3 } \\
{ ( u + v ) ( u ^ { 2 } + v ^ { 2 } - u v ) = 9 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ ( u + v ) ^ { 2 } - 3 u v = 3 , } \\
{ u + v = 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
u v=2, \\
u+v=3
\end{array} \Leftrightarrow\right.\right.\right. \\
& \Leftrightarrow\left[\begin{array} { l } 
{ u = 2 , v = 1 , } \\
{ u = 1 , v = 2 }
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=1 \\
x=-6
\end{array}\right.\right.
\end{aligned}
$$

Answer: $x_{1}=1, x_{2}=-6$.