6.218. $\left\{\begin{array}{l}x^{3}+y^{3}=9 a^{3}, \\ x^{2} y+x y^{2}=6 a^{3} .\end{array}\right.$
## Solution.

Let's rewrite the system of equations as

$$
\left\{\begin{array}{l}
(x+y)\left((x+y)^{2}-3 x y\right)=9 a^{3} \\
x y(x+y)=6 a^{3}
\end{array}\right.
$$

Let $\left\{\begin{array}{l}x+y=u, \\ x y=v .\end{array}\right.$ Then, in terms of $u$ and $v$, the system becomes

$$
\left\{\begin{array}{l}
u\left(u^{2}-3 v\right)=9 a^{3}, \\
u v=6 a^{3}
\end{array} \Rightarrow v=\frac{6 a^{3}}{u}, u\left(u^{2}-3 \cdot \frac{6 a^{3}}{u}\right)=9 a^{3} \Rightarrow u^{3}=27 a^{3}\right.
$$

from which $u=3 a ; v=\frac{6 a^{3}}{3 a}=2 a^{2}$.

Then $\left\{\begin{array}{l}x+y=3 a, \\ x y=2 a^{2}\end{array} \Leftrightarrow\left\{\begin{array}{l}x_{1}=a, \\ y_{1}=2 a ;\end{array}\left\{\begin{array}{l}x_{2}=2 a, \\ y_{2}=a .\end{array}\right.\right.\right.$

Answer: $(a ; 2 a),(2 a ; a)$.