418. Solve the system of equations:

$$
\left\{\begin{array}{l}
x^{2}+x \sqrt[3]{x y^{2}}=80 \\
y^{2}+y \sqrt[3]{x^{2} y}=5
\end{array}\right.
$$
$\triangle$ Let $\frac{x}{y}=t$, i.e., $y=t x$, considering that from the first equation $x \neq 0$. We will have:

$$
\begin{aligned}
& x^{2}+x^{2 \sqrt[3]{t^{2}}}=80, \quad t^{2} x^{2}+t x^{23} \sqrt{t^{2}}=5 \\
& x^{2}\left(1+\sqrt[3]{t^{2}}\right)=80, \quad x^{2} t^{\frac{4}{3}}\left(1+\sqrt[3]{t^{2}}\right)=5
\end{aligned}
$$

Divide the second equation of the last system by the first:

$$
t^{\frac{4}{3}}=\frac{1}{16}, \quad t^{4}=\frac{1}{2^{12}}, \quad t= \pm \frac{1}{8}
$$

1) Let $t=\frac{1}{8}$. Then $\frac{y}{x}=\frac{1}{8}, x=8 y$. From the first equation of the last system, we find $y$:

$$
64 y^{2} \cdot \frac{5}{4}=80, \quad y^{2}=1, \quad y= \pm 1
$$

If $y=1$, then $x=8$; if $y=-1$, then $x=-8$.

2) Let $t=-\frac{1}{8}$. Similarly, we obtain the solutions $(8 ;-1)$ and $(-8 ; 1)$.

Verification shows that all found solutions satisfy the original system of equations.

Answer: ( $\pm 8 ; \pm 1$ ) (four solutions).