330. Solve the system of equations:

$$
\left\{\begin{array}{l}
x^{3}+y^{3}=7(x+y) \\
x^{3}-y^{3}=13(x-y)
\end{array}\right.
$$
$\triangle$ Let's gather all terms in each equation of the system to the left side and factor out $(x+y)$ in the first equation and $-x-y$ in the second:

$$(x+y)\left(x^{2}-x y+y^{2}-7\right)=0, \quad(x-y)\left(x^{2}+x y+y^{2}-13\right)=0.$$

When solving the last system of equations, we need to consider four cases.

1) Suppose the first factors in the left parts of both equations are zero:

$$
x+y=0, \quad x-y=0
$$

We get the solution $(0 ; 0)$.

2) Suppose the first factor from the first equation and the second factor from the second equation are zero:

$$
x+y=0, \quad x^{2}+x y+y^{2}-13=0
$$

From this, $x^{2}=13$, i.e., $x= \pm \sqrt{13}$. We have the solutions $(\sqrt{13} ;-\sqrt{13})$ and $(-\sqrt{13} ; \sqrt{13})$.

3) Suppose

$$
x-y=0, \quad x^{2}-x y+y^{2}-7=0
$$

We find the solutions $(\sqrt{7} ; \sqrt{7})$ and $(-\sqrt{7} ;-\sqrt{7})$.

4) Suppose

$$
x^{2}-x y+y^{2}-7=0, \quad x^{2}+x y+y^{2}-13=0
$$

i.e.,

$$
x^{2}+x y+y^{2}=13, \quad x^{2}-x y+y^{2}=7
$$

Adding and subtracting these equations, we find $x^{2}+y^{2}$ and $2 x y$:

$$
x^{2}+y^{2}=10, \quad 2 x y=6
$$

Adding and subtracting the last equations:

$$
(x+y)^{2}=16, \quad(x-y)^{2}=4
$$

i.e.,

$$
x+y= \pm 4, \quad x-y= \pm 2
$$

It remains to consider four sub-cases, depending on the signs in the right parts of the last equations. Complete the solution independently.

Answer: $(0 ; 0),(3 ; 1),(1 ; 3),(-3 ;-1),(-1 ;-3),(\sqrt{13} ;-\sqrt{13}),(-\sqrt{13} ; \sqrt{13})$, $(\sqrt{7} ; \sqrt{7}),(-\sqrt{7} ;-\sqrt{7})$.