6.208. $\left\{\begin{array}{l}(x-y)\left(x^{2}-y^{2}\right)=3 a^{3}, \\ (x+y)\left(x^{2}+y^{2}\right)=15 a^{3} .\end{array}\right.$

The system of equations is:
\[
\left\{\begin{array}{l}
(x-y)\left(x^{2}-y^{2}\right)=3 a^{3}, \\
(x+y)\left(x^{2}+y^{2}\right)=15 a^{3} .
\end{array}\right.
\]
Solution.

Rewrite the system of equations as

$$
\left\{\begin{array}{l}
x^{3}-x y^{2}-x^{2} y+y^{3}=3 a^{3} \\
x^{2}+x y^{2}+x^{2} y+y^{3}=15 a^{3}
\end{array}\right.
$$

Add and subtract the second and first equations:

$$
\left\{\begin{array} { l } 
{ 2 x ^ { 3 } + 2 y ^ { 3 } = 1 8 a ^ { 3 } } \\
{ 2 x y ^ { 2 } + 2 x ^ { 2 } y = 1 2 a ^ { 3 } }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x ^ { 3 } + y ^ { 3 } = 9 a ^ { 3 } } \\
{ x y ( x + y ) = 6 a ^ { 3 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x+y)\left((x+y)^{2}-3 x y\right)=9 a^{3} \\
x y(x+y)=6 a^{3}
\end{array}\right.\right.\right.
$$

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

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

$$
u=3 a, v=\frac{6 a^{3}}{3 a}=2 a^{2} \Rightarrow\left\{\begin{array} { l } 
{ x + y = 3 a , } \\
{ x y = 2 a ^ { 2 } }
\end{array} \Rightarrow \left\{\begin{array}{l}
x_{1}=a, \\
y_{1}=2 a
\end{array} ;\left\{\begin{array}{l}
x_{2}=2 a \\
y_{2}=a
\end{array}\right.\right.\right.
$$

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