342. Solve the system of equations:

$$
\left\{\begin{array}{l}
x^{2}+y=13 \\
x^{6}+y^{3}=793
\end{array}\right.
$$
$\triangle$ The given system is not symmetric. Let $x^{2}=z$. Then we obtain a symmetric system:

$$
\left\{\begin{array}{l}
z+y=13 \\
z^{3}+y^{3}=793
\end{array}\right.
$$

Introduce the elementary symmetric polynomials $u=z+y, v=zy$. We will have:

$$
u=13, \quad u^{3}-3uv=793
$$

From this, $u=13, v=36$, i.e.,

$$
z+y=13, \quad zy=36
$$

Solving this system of equations, we get:

$$
z_{1}=9, \quad y_{1}=4 ; \quad z_{2}=4, \quad y_{2}=9
$$

It remains to find $x$.

Answer: $(3 ; 4),(-3 ; 4),(2 ; 9),(-2 ; 9)$.