6.188. $\left\{\begin{array}{l}x-a y+a^{2} z=a^{3}, \\ x-b y+b^{2} z=b^{3}, \\ x-c y+c^{2} z=c^{3}, \quad a \neq b, b \neq c, c \neq a .\end{array}\right.$
## Solution.

Subtracting the second equation of the system from the first, we have

$$
\begin{aligned}
& -(a-b) y+(a-b)(a+b) z=(a-b)\left(a^{2}+a b+b^{2}\right) \Leftrightarrow \\
& \Leftrightarrow-y+(a+b) z=a^{2}+a b+b^{2}
\end{aligned}
$$

Similarly, subtracting the third equation from the first, we get

$$
\begin{aligned}
& -(a-c) y+(a-c)(a+c) z=(a-c)\left(a^{2}+a c+c^{2}\right) \Leftrightarrow \\
& \Leftrightarrow-y+(a+c) z=a^{2}+a c+c^{2}
\end{aligned}
$$

Subtracting equation (2) from equation (1), we have

$$
(a+b) z-(a+c) z=a b+b^{2}-a c-c^{2} \Leftrightarrow z=a+b+c
$$

Substituting \( z=a+b+c \) into equation (1), we get \( y=a b+b c+c a \). Substituting \( z=a+b+c \) and \( y=a b+b c+c a \) into the first equation of the system, we have \( x=a b c \).

Answer: \( x=a b c, y=a b+b c+c a, z=a+b+c \).