7. The equation
$$
3 x^{3}+2 \sqrt{2} x^{2}-(17-9 \sqrt{2}) x-(6-5 \sqrt{2})=0
$$

has solutions $x_{1}=$ $\qquad$ ,$x_{2}=$ $\qquad$ ,$x_{3}=$ $\qquad$
$7 \cdot \frac{\sqrt{2}}{3}, \sqrt{2}-1,1-2 \sqrt{2}$.
Let $x=\sqrt{2} y$, substituting into the original equation we get
$$
6 \sqrt{2} y^{3}+4 \sqrt{2} y^{2}-17 \sqrt{2} y+18 y-6+5 \sqrt{2}=0 \text {. }
$$

It is easy to see that $y=\frac{1}{3}$ satisfies the condition. Therefore, $x_{1}=\frac{\sqrt{2}}{3}$. Thus,
$$
\begin{array}{l}
3 x^{3}+2 \sqrt{2} x^{2}-(17-9 \sqrt{2}) x-(6-5 \sqrt{2}) \\
=\left(x-\frac{\sqrt{2}}{3}\right)\left(3 x^{2}+3 \sqrt{2} x+9 \sqrt{2}-15\right) . \\
=3\left(x-\frac{\sqrt{2}}{3}\right)(x-\sqrt{2}+1)(x+2 \sqrt{2}-1) .
\end{array}
$$

Therefore, $x_{1}=\frac{\sqrt{2}}{3}, x_{2}=\sqrt{2}-1, x_{3}=1-2 \sqrt{2}$.