Solve the following system of equations:

$$
\begin{gathered}
\frac{x-1}{y-1}=\frac{a-1}{b-1} \\
\frac{x^{3}-1}{y^{3}-1}=\frac{a^{3}-1}{b^{3}-1}
\end{gathered}
$$
I. Solution: We must assume that $b \neq 1$ and exclude the value $y=1$, because otherwise our system of equations would make no sense; furthermore, we can assume that $a \neq 1$, because otherwise $x=1$ and $y$ could be any value. Thus, $x \neq 1$.

Divide equation (2) by (1):

$$
\frac{\left(x^{3}-1\right)(y-1)}{\left(y^{3}-1\right)(x-1)}=\frac{\left(a^{3}-1\right)(b-1)}{\left(b^{3}-1\right)(a-1)}
$$

From this,

$$
\frac{x^{2}+x+1}{y^{2}+y+1}=\frac{a^{2}+a+1}{b^{2}+b+1}
$$

or

$$
\left(b^{2}+b+1\right)\left(x^{2}+x+1\right)=\left(a^{2}+a+1\right)\left(y^{2}+y+1\right)
$$

From (1),

$$
x=\frac{a-1}{b-1}(y-1)+1
$$

and thus

$$
x^{2}+x+1=x(x+1)+1=\frac{(a-1)^{2}}{(b-1)^{2}}\left(y^{2}-2 y+1\right)+\frac{3(a-1)}{b-1}(y-1)+3
$$

Substituting this value into (3) and multiplying both sides by $(b-1)^{2}$,

$$
\begin{gathered}
\left(b^{2}+b+1\right)\left[(a-1)^{2}\left(y^{2}-2 y-1\right)+3(a-1)(b-1)(y-1)+3(b-1)^{2}\right]= \\
=(b-1)^{2}\left(a^{2}+a+1\right)\left(y^{2}+y+1\right)
\end{gathered}
$$

Rearranging by powers of $y$,

$$
\begin{aligned}
\left(a^{2} b-a b^{2}+b-a\right) y^{2} & +\left(a b^{3}-a^{2} b^{2}+2 a b-a^{2}-b^{3}+a b^{2}-b^{2}\right) y+ \\
& +(b-a)\left(b^{2}-a\right) b=0
\end{aligned}
$$

or

$$
(b-a)(1-a b) y^{2}+(b-a)\left[b(a b-1)+a-b^{2}\right] y+(b-a)\left(b^{2}-a\right) b=0
$$

If $a=b$, our equation becomes an identity, and every $x=y$ value is a solution to our system. 

If $a b=1$, then $b=\frac{1}{a}$, and thus the right side of (1) is $-a$, and the right side of (2) is $-a^{3}=(-a)^{3}$, which leads to a contradiction.

Therefore, we can assume that $a-b \neq 0$ and $1-a b \neq 0$.

Dividing by $(b-a)(1-a b)$, we get

$$
y^{2}-\left(b+\frac{b^{2}-a}{1-a b}\right) y+\frac{b\left(b^{2}-a\right)}{1-a b}=0
$$

from which

$$
y_{1}=b, \quad y_{2}=\frac{b^{2}-a}{1-a b}
$$

and thus from (4),

$$
x_{1}=a, \quad x_{2}=\frac{a^{2}-b}{1-a b}
$$

The trivial solution $x_{1}=a, y_{1}=b$ is otherwise directly visible.

Remark: Each of the roots $x_{2}, y_{2}$, as we have seen, must differ from 1, so $a^{2}-b \neq 1-a b$, or $(a-1)(a+1+b) \neq 0$. Thus, $x_{2} \neq 1$ if $a+b+1 \neq 0$. The same condition ensures that $y_{2} \neq 1$.

$x_{2}, y_{2}$ differ from the trivial solution only if $\frac{a^{2}-b}{1-a b} \neq a$, or $(a-1)(a b+b+a) \neq 0$, i.e., if $a b+b+a \neq 0$. At the same time, $y_{2} \neq b$.

II. Solution. We can directly see the trivial solution $x_{1}=a, y_{1}=b$. Just as in the first solution, from dividing (2) by (1), we get

$$
\frac{x^{2}+x+1}{a^{2}+a+1}=\frac{y^{2}+y+1}{b^{2}+b+1}
$$

Let, based on (1),

$$
\frac{x-1}{a-1}=\frac{y-1}{b-1}=z
$$

From this,

$$
x=(a-1) z+1, \quad y=(b-1) z+1
$$

Substituting these values into (3),

$$
\frac{[(a-1) z+1]^{2}+(a-1) z+2}{a^{2}+a+1}=\frac{[(b-1) z+1]^{2}+(b-1) z+2}{b^{2}+b+1}
$$

Removing the fraction, rearranging, and simplifying by $3(b-a)$,

$$
(1-a b) z^{2}+(a b-a-b-2) z+(a+b+1)=0
$$

The trivial solution $x_{1}=a, y_{1}=b$ provides the value $z=1$, so $z=1$ is one root of equation (4). Dividing by the $(z-1)$ root factor,

$$
(1-a b) z-(a+b+1)=0
$$

from which

$$
z=\frac{a+b+1}{1-a b}=\frac{x-1}{a-1}=\frac{y-1}{b-1}
$$

From this,

$$
x_{2}=\frac{a^{2}-b}{1-a b}, \quad y_{2}=\frac{b^{2}-a}{1-a b}
$$

Vigassy György (Bp. I., Petófi g. IV. o.t.)