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

We obtain the solution of the system using Cramer's rule. Let's compute the determinants of the system

$$
\Delta=\left|\begin{array}{ccc}
a & b & c \\
a^{2} & b^{2} & c^{2} \\
a^{3} & b^{3} & c^{3}
\end{array}\right|=a b^{2} c^{3}+a^{3} b c^{2}+a^{2} b^{3} c-a^{3} b^{2} c-a^{2} b c^{3}-a b^{3} c^{2}
$$

$$
\begin{aligned}
& \Delta x=\left|\begin{array}{ccc}
k & b & c \\
k^{2} & b^{2} & c^{2} \\
k^{3} & b^{3} & c^{3}
\end{array}\right|=k b^{2} c^{3}+k^{3} b c^{2}+k^{2} b^{3} c-k^{3} b^{2} c-k^{2} b c^{3}-k b^{3} c^{2} \\
& \Delta y=\left|\begin{array}{ccc}
a & k & c \\
a^{2} & k^{2} & c^{2} \\
a^{3} & k^{3} & c^{3}
\end{array}\right|=k^{2} a c^{3}+k^{3} a^{2} c+k a^{3} c^{2}-k^{2} a^{3} c-k a^{2} c^{3}-k^{3} a c^{2} \\
& \Delta z=\left|\begin{array}{ccc}
a & b & k \\
a^{2} & b^{2} & k^{2} \\
a^{3} & b^{3} & k^{3}
\end{array}\right|=k^{3} a b^{2}+k^{2} a^{3} b+k a^{2} b^{3}-k a^{3} b^{2}-k^{2} a b^{3}-k^{3} a^{2} b
\end{aligned}
$$

The solution of the system is

$$
\begin{aligned}
& x=\frac{\Delta x}{\Delta}=\frac{k b^{2} c^{3}+k^{3} b c^{2}+k^{2} b^{3} c-k^{3} b^{2} c-k^{2} b c^{3}-k b^{3} c^{2}}{a b^{2} c^{3}+a^{3} b c^{2}+a^{2} b^{3} c-a^{3} b^{2} c-a^{2} b c^{3}-a b^{3} c^{2}} \\
& x=\frac{k(k-c)(k-b)}{a(a-c)(a-b)} \\
& y=\frac{\Delta y}{\Delta}=\frac{k^{2} a c^{3}+k^{3} a^{2} c+k a^{3} c^{2}-k^{2} a^{3} c-k a^{2} c^{3}-k^{3} a c^{2}}{a b^{2} c^{3}+a^{3} b c^{2}+a^{2} b^{3} c-a^{3} b^{2} c-a^{2} b c^{3}-a b^{3} c^{2}} \\
& y=\frac{k(k-c)(k-a)}{b(b-c)(b-a)}
\end{aligned}
$$

$$
\begin{aligned}
& z=\frac{\Delta z}{\Delta}=\frac{k^{3} a b^{2}+k^{2} a^{3} b+k a^{2} b^{3}-k a^{3} b^{2}-k^{2} a b^{3}-k^{3} a^{2} b}{a b^{2} c^{3}+a^{3} b c^{2}+a^{2} b^{3} c-a^{3} b^{2} c-a^{2} b c^{3}-a b^{3} c^{2}} \\
& z=\frac{k(k-a)(k-b)}{c(c-a)(c-b)}
\end{aligned}
$$

Answer: \( x=\frac{k(k-c)(k-b)}{a(a-c)(a-b)} ; y=\frac{k(k-c)(k-a)}{b(b-c)(b-a)} ; z=\frac{k(k-a)(k-b)}{c(c-a)(c-b)} \).