$6.298 \frac{\left(\sqrt[3]{(15-x)^{2}}+\sqrt[3]{(15-x)(x-6)}+\sqrt[3]{(x-6)^{2}}\right)^{2}}{\sqrt[3]{15-x}+\sqrt[3]{x-6}}=\frac{49}{3}$.
Solution. Let $u=\sqrt[3]{15-x}, v=\sqrt[3]{x-6}$, then

$$
\left\{\begin{array} { l } 
{ \frac { ( u ^ { 2 } + u v + v ^ { 2 } ) ^ { 2 } } { u + v } = \frac { 4 9 } { 3 } , } \\
{ u ^ { 3 } + v ^ { 3 } = 9 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{\left.(u+v)^{2}-u v\right)^{2}}{u+v}=\frac{49}{3} \\
(u+v)\left((u+v)^{2}-3 u v\right)=9
\end{array}\right.\right.
$$

Now let $y=u+v, z=u v$, then

$$
\left\{\begin{array} { l } 
{ \frac { ( y ^ { 2 } - z ) ^ { 2 } } { y } = \frac { 4 9 } { 3 } , } \\
{ y ( y ^ { 2 } - 3 z ) = 9 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ ( y ^ { 2 } - \frac { y ^ { 3 } - 9 } { 3 y } ) ^ { 2 } = \frac { 4 9 } { 3 } y , } \\
{ z = \frac { y ^ { 3 } - 9 } { 3 y } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\left(2 y^{3}+9\right)^{2}=147 y^{3}, \\
z=\frac{y^{3}-9}{3 y}
\end{array} \Rightarrow\right.\right.\right.
$$

$$
\begin{aligned}
& \Rightarrow 4 y^{6}-111 y^{3}+81=0 \Leftrightarrow\left[\begin{array} { l } 
{ y ^ { 3 } = 2 7 , } \\
{ y ^ { 3 } = \frac { 3 } { 4 } }
\end{array} \Rightarrow \left[\begin{array}{l}
z=2, \\
z=-\frac{11}{\sqrt[3]{48}}
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left[\begin{array} { l } 
{ z ^ { 3 } = 8 , } \\
{ z ^ { 3 } = - \frac { 1 3 3 1 } { 4 8 } }
\end{array} \Leftrightarrow \left[\begin{array}{l}
(15-x)(x-6)=8 \\
(15-x)(x-6)=-\frac{1331}{48}
\end{array}\right.\right.
\end{aligned}
$$

1) $x^{2}-21 x+98=0 \Leftrightarrow\left[\begin{array}{l}x=7 \\ x=14 .\end{array}\right.$
2) $x^{2}-21 x+90-\frac{1331}{48}=0 \Leftrightarrow x=\frac{21}{2} \pm \sqrt{\frac{49 \cdot 47}{48}}=\frac{21}{2} \pm \frac{7 \sqrt{141}}{12}$.

Answer: $x_{1}=7, x_{2}=14, x_{3,4}=\frac{21}{2} \pm \frac{7 \sqrt{141}}{12}$.