Find all real solutions of the system \[\begin{cases}x^2 + y^2 - z(x + y) = 2, \\ y^2 + z^2 - x(y + z) = 4, \\ z^2 + x^2 - y(z + x) = 8.\end{cases}\]
1. Start with the given system of equations:
   \[
   \begin{cases}
   x^2 + y^2 - z(x + y) = 2, \\
   y^2 + z^2 - x(y + z) = 4, \\
   z^2 + x^2 - y(z + x) = 8.
   \end{cases}
   \]

2. Subtract the second equation from the first:
   \[
   (x^2 + y^2 - z(x + y)) - (y^2 + z^2 - x(y + z)) = 2 - 4,
   \]
   which simplifies to:
   \[
   x^2 - z^2 - z(x + y) + x(y + z) = -2.
   \]
   Rearrange and factor:
   \[
   (x - z)(x + z) + x(y + z) - z(x + y) = -2,
   \]
   \[
   (x - z)(x + y + z) = -2. \quad \text{(Equation 4)}
   \]

3. Subtract the third equation from the second:
   \[
   (y^2 + z^2 - x(y + z)) - (z^2 + x^2 - y(z + x)) = 4 - 8,
   \]
   which simplifies to:
   \[
   y^2 - x^2 - x(y + z) + y(z + x) = -4.
   \]
   Rearrange and factor:
   \[
   (y - x)(y + x) + y(z + x) - x(y + z) = -4,
   \]
   \[
   (y - x)(x + y + z) = -4. \quad \text{(Equation 5)}
   \]

4. Subtract the first equation from the third:
   \[
   (z^2 + x^2 - y(z + x)) - (x^2 + y^2 - z(x + y)) = 8 - 2,
   \]
   which simplifies to:
   \[
   z^2 - y^2 - y(z + x) + z(x + y) = 6.
   \]
   Rearrange and factor:
   \[
   (z - y)(z + y) + z(x + y) - y(z + x) = 6,
   \]
   \[
   (z - y)(x + y + z) = 6. \quad \text{(Equation 6)}
   \]

5. Add the three equations:
   \[
   (x - z)^2 + (y - x)^2 + (z - y)^2 = 14.
   \]

6. Let \( x + y + z = a \). Then from Equations 4, 5, and 6:
   \[
   \left(\frac{-2}{a}\right)^2 + \left(\frac{-4}{a}\right)^2 + \left(\frac{6}{a}\right)^2 = 14,
   \]
   \[
   \frac{4}{a^2} + \frac{16}{a^2} + \frac{36}{a^2} = 14,
   \]
   \[
   \frac{56}{a^2} = 14,
   \]
   \[
   a^2 = 4,
   \]
   \[
   a = \pm 2.
   \]

7. Substitute \( a = \pm 2 \) back into Equations 4, 5, and 6:
   - For \( a = 2 \):
     \[
     x - z = -1, \quad y - x = -2, \quad z - y = 3.
     \]
     Solving these:
     \[
     x - z = -1 \implies x = z - 1,
     \]
     \[
     y - x = -2 \implies y = x - 2,
     \]
     \[
     z - y = 3 \implies z = y + 3.
     \]
     Substitute \( y = x - 2 \) into \( z = y + 3 \):
     \[
     z = (x - 2) + 3 = x + 1.
     \]
     Substitute \( z = x + 1 \) into \( x = z - 1 \):
     \[
     x = (x + 1) - 1 \implies x = x.
     \]
     Thus, \( x = 1, y = -1, z = 2 \).

   - For \( a = -2 \):
     \[
     x - z = 1, \quad y - x = 2, \quad z - y = -3.
     \]
     Solving these:
     \[
     x - z = 1 \implies x = z + 1,
     \]
     \[
     y - x = 2 \implies y = x + 2,
     \]
     \[
     z - y = -3 \implies z = y - 3.
     \]
     Substitute \( y = x + 2 \) into \( z = y - 3 \):
     \[
     z = (x + 2) - 3 = x - 1.
     \]
     Substitute \( z = x - 1 \) into \( x = z + 1 \):
     \[
     x = (x - 1) + 1 \implies x = x.
     \]
     Thus, \( x = -1, y = 1, z = -2 \).

Conclusion:
The solutions are \((1, -1, 2)\) and \((-1, 1, -2)\).

The final answer is \( \boxed{ (1, -1, 2) } \) and \((-1, 1, -2)\).