Solve the system of equations in real numbers:
\[
\begin{cases*}
(x - 1)(y - 1)(z - 1) = xyz - 1,\\
(x - 2)(y - 2)(z - 2) = xyz - 2.\\
\end{cases*}
\]
[i]Vladimir Bragin[/i]
1. **Define the polynomial \( P(t) \):**
   Let \( P(t) = (t - x)(t - y)(t - z) = t^3 - at^2 + bt - c \).

2. **Evaluate \( P(t) \) at specific points:**
   Given the system of equations:
   \[
   \begin{cases*}
   (x - 1)(y - 1)(z - 1) = xyz - 1, \\
   (x - 2)(y - 2)(z - 2) = xyz - 2,
   \end{cases*}
   \]
   we can substitute \( t = 1 \) and \( t = 2 \) into \( P(t) \):
   \[
   P(1) = (1 - x)(1 - y)(1 - z) = 1 - c,
   \]
   \[
   P(2) = (2 - x)(2 - y)(2 - z) = 8 - 4a + 2b - c.
   \]

3. **Use the given equations:**
   From the first equation:
   \[
   (x - 1)(y - 1)(z - 1) = xyz - 1 \implies 1 - c = xyz - 1 \implies c = xyz.
   \]
   From the second equation:
   \[
   (x - 2)(y - 2)(z - 2) = xyz - 2 \implies 8 - 4a + 2b - c = xyz - 2.
   \]

4. **Relate the coefficients:**
   Since \( c = xyz \), we substitute \( c \) in the second equation:
   \[
   8 - 4a + 2b - xyz = xyz - 2.
   \]
   Simplifying, we get:
   \[
   8 - 4a + 2b = 2xyz - 2.
   \]
   Since \( c = xyz \), we have:
   \[
   8 - 4a + 2b = 2c - 2.
   \]

5. **Solve for \( a \) and \( b \):**
   From the polynomial \( P(t) = t^3 - at^2 + bt - c \), we know:
   \[
   P(1) = 1 - a + b - c = 1 - c,
   \]
   \[
   P(2) = 8 - 4a + 2b - c = 2 - c.
   \]
   Substituting \( c = 1 \) (since \( c = xyz \) and we are looking for real roots):
   \[
   1 - a + b - 1 = 0 \implies -a + b = 0 \implies a = b,
   \]
   \[
   8 - 4a + 2a - 1 = 1 \implies 8 - 2a - 1 = 1 \implies 7 - 2a = 1 \implies 2a = 6 \implies a = 3.
   \]
   Therefore, \( a = b = 3 \).

6. **Form the polynomial with the found coefficients:**
   \[
   P(t) = t^3 - 3t^2 + 3t - 1.
   \]

7. **Find the roots of \( P(t) \):**
   The polynomial \( P(t) = (t - 1)^3 \) has a triple root at \( t = 1 \).

8. **Conclude the solution:**
   Since \( P(t) = (t - 1)^3 \), the roots are \( x = y = z = 1 \).

The final answer is \( \boxed{x = y = z = 1} \).