$6.3016 \sqrt[3]{x-3}+\sqrt[3]{x-2}=5 \sqrt[6]{(x-2)(x-3)}$.
Solution. There are two cases:

54

1) $\left\{\begin{array}{l}x-2 \geq 0, \\ x-3 \geq 0,\end{array} \Rightarrow x \geq 3\right.$. Let $u=\sqrt[6]{x-2}, v=\sqrt[6]{x-3}$, then

$6 v^{2}+u^{2}=5 u v \Leftrightarrow u^{2}-5 u v+6 v^{2}=0 \Leftrightarrow u=\frac{5 v \pm \sqrt{25 v^{2}-24 v^{2}}}{2} \Leftrightarrow$

$\Leftrightarrow\left[\begin{array}{l}u=3 v, \\ u=2 v\end{array} \Leftrightarrow\left[\begin{array}{l}x-2=3^{6} \cdot(x-3), \\ x-2=2^{6} \cdot(x-3)\end{array} \Leftrightarrow\left[\begin{array}{l}x=\frac{2185}{728}, \\ x=\frac{190}{63} .\end{array}\right.\right.\right.$

2) $\left\{\begin{array}{l}x-2 \leq 0, \\ x-3 \leq 0\end{array} \Rightarrow x \leq 2\right.$ and in this case the left side of the original equation is negative, while the right side is positive.

Answer: $x_{1}=\frac{190}{63}, x_{2}=\frac{2185}{728}$.

## $6.302 x^{3}+x+\sqrt[3]{x^{3}+x-2}=12$

Solution. $x^{3}+x-2+\sqrt[3]{x^{3}+x-2}=10$. Let $y=\sqrt[3]{x^{3}+x-2}$, then $y^{3}+y-10=0 \Leftrightarrow(y-2)\left(y^{2}+2 y+5\right)=0 \Leftrightarrow y=2 \Leftrightarrow$ $\Leftrightarrow x^{3}+x-10=0 \Leftrightarrow(x-2)\left(x^{2}+2 x+5\right)=0 \Leftrightarrow x=2$.

Answer: $x=2$.

## Solve the systems of equations ( $6.303-6.341$ ):

$6.303\left\{\begin{array}{l}(x+y)\left(x^{2}-y^{2}\right)=16 \\ (x-y)\left(x^{2}+y^{2}\right)=40 .\end{array}\right.$

Solution. Divide the equations term by term:

$$
\begin{aligned}
& \frac{(x+y)(x-y)(x+y)}{(x-y)\left(x^{2}+y^{2}\right)}=\frac{2}{5} \Rightarrow 3 x^{2}+10 x y+3 y^{2}=0 \Leftrightarrow \\
& \Leftrightarrow x=\frac{-10 y \pm \sqrt{100 y^{2}-36 y^{2}}}{6}=\frac{-10 y \pm 8 y}{6} \Leftrightarrow\left[\begin{array}{l}
x=-3 y, \\
x=-\frac{1}{3} y .
\end{array}\right.
\end{aligned}
$$

1) $\left\{\begin{array}{l}x=-3 y, \\ (x+y)\left(x^{2}-y^{2}\right)=16\end{array} \Leftrightarrow\left\{\begin{array}{l}x=-3 y, \\ y^{3}=-1\end{array} \Leftrightarrow\left\{\begin{array}{l}x=3, \\ y=-1 .\end{array}\right.\right.\right.$
2) $\left\{\begin{array}{l}y=-3 x, \\ (x+y)\left(x^{2}-y^{2}\right)=16\end{array} \Leftrightarrow\left\{\begin{array}{l}y=-3 x, \\ x^{3}=1\end{array} \Leftrightarrow\left\{\begin{array}{